Given a sorted array of n uniformly distributed values arr[], write a function to search for a particular element x in the array.
Linear Search finds the element in O(n) time, Jump Search takes O(n) time and Binary Search takes O(log n) time.
The Interpolation Search is an improvement over Binary Search for instances, where the values in a sorted array are uniformly distributed. Interpolation constructs new data points within the range of a discrete set of known data points. Binary Search always goes to the middle element to check. On the other hand, interpolation search may go to different locations according to the value of the key being searched. For example, if the value of the key is closer to the last element, interpolation search is likely to start search toward the end side.
To find the position to be searched, it uses the following formula.
// The idea of formula is to return higher value of pos
// when element to be searched is closer to arr[hi]. And
// smaller value when closer to arr[lo]
arr[] ==> Array where elements need to be searched
x ==> Element to be searched
lo ==> Starting index in arr[]
hi ==> Ending index in arr[]
pos = lo + [ \frac{(x-arr[lo])*(hi-lo) }{ (arr[hi]-arr[Lo]) }]
There are many different interpolation methods and one such is known as linear interpolation. Linear interpolation takes two data points which we assume as (x1,y1) and (x2,y2) and the formula is : at point(x,y).
This algorithm works in a way we search for a word in a dictionary. The interpolation search algorithm improves the binary search algorithm. The formula for finding a value is: K = data-low/high-low.
K is a constant which is used to narrow the search space. In the case of binary search, the value for this constant is: K=(low+high)/2.
The formula for pos can be derived as follows.
Let's assume that the elements of the array are linearly distributed.
General equation of line : y = m*x + c.
y is the value in the array and x is its index.
Now putting value of lo,hi and x in the equation
arr[hi] = m*hi+c ----(1)
arr[lo] = m*lo+c ----(2)
x = m*pos + c ----(3)
m = (arr[hi] - arr[lo] )/ (hi - lo)
subtracting eqxn (2) from (3)
x - arr[lo] = m * (pos - lo)
lo + (x - arr[lo])/m = pos
pos = lo + (x - arr[lo]) *(hi - lo)/(arr[hi] - arr[lo])
Algorithm
The rest of the Interpolation algorithm is the same except for the above partition logic.
- Step1: In a loop, calculate the value of "pos" using the probe position formula.
- Step2: If it is a match, return the index of the item, and exit.
- Step3: If the item is less than arr[pos], calculate the probe position of the left sub-array. Otherwise, calculate the same in the right sub-array.
- Step4: Repeat until a match is found or the sub-array reduces to zero.
Below is the implementation of the algorithm.
C++
// C++ program to implement interpolation
// search with recursion
#include <bits/stdc++.h>
using namespace std;
// If x is present in arr[0..n-1], then returns
// index of it, else returns -1.
int interpolationSearch(int arr[], int lo, int hi, int x)
{
int pos;
// Since array is sorted, an element present
// in array must be in range defined by corner
if (lo <= hi && x >= arr[lo] && x <= arr[hi]) {
// Probing the position with keeping
// uniform distribution in mind.
pos = lo
+ (((double)(hi - lo) / (arr[hi] - arr[lo]))
* (x - arr[lo]));
// Condition of target found
if (arr[pos] == x)
return pos;
// If x is larger, x is in right sub array
if (arr[pos] < x)
return interpolationSearch(arr, pos + 1, hi, x);
// If x is smaller, x is in left sub array
if (arr[pos] > x)
return interpolationSearch(arr, lo, pos - 1, x);
}
return -1;
}
// Driver Code
int main()
{
// Array of items on which search will
// be conducted.
int arr[] = { 10, 12, 13, 16, 18, 19, 20, 21,
22, 23, 24, 33, 35, 42, 47 };
int n = sizeof(arr) / sizeof(arr[0]);
// Element to be searched
int x = 18;
int index = interpolationSearch(arr, 0, n - 1, x);
// If element was found
if (index != -1)
cout << "Element found at index " << index;
else
cout << "Element not found.";
return 0;
}
// This code is contributed by equbalzeeshan
// C program to implement interpolation search
// with recursion
#include <stdio.h>
// If x is present in arr[0..n-1], then returns
// index of it, else returns -1.
int interpolationSearch(int arr[], int lo, int hi, int x)
{
int pos;
// Since array is sorted, an element present
// in array must be in range defined by corner
if (lo <= hi && x >= arr[lo] && x <= arr[hi]) {
// Probing the position with keeping
// uniform distribution in mind.
pos = lo
+ (((double)(hi - lo) / (arr[hi] - arr[lo]))
* (x - arr[lo]));
// Condition of target found
if (arr[pos] == x)
return pos;
// If x is larger, x is in right sub array
if (arr[pos] < x)
return interpolationSearch(arr, pos + 1, hi, x);
// If x is smaller, x is in left sub array
if (arr[pos] > x)
return interpolationSearch(arr, lo, pos - 1, x);
}
return -1;
}
// Driver Code
int main()
{
// Array of items on which search will
// be conducted.
int arr[] = { 10, 12, 13, 16, 18, 19, 20, 21,
22, 23, 24, 33, 35, 42, 47 };
int n = sizeof(arr) / sizeof(arr[0]);
int x = 18; // Element to be searched
int index = interpolationSearch(arr, 0, n - 1, x);
// If element was found
if (index != -1)
printf("Element found at index %d", index);
else
printf("Element not found.");
return 0;
}
// Java program to implement interpolation
// search with recursion
import java.util.*;
class GFG {
// If x is present in arr[0..n-1], then returns
// index of it, else returns -1.
public static int interpolationSearch(int arr[], int lo,
int hi, int x)
{
int pos;
// Since array is sorted, an element
// present in array must be in range
// defined by corner
if (lo <= hi && x >= arr[lo] && x <= arr[hi]) {
// Probing the position with keeping
// uniform distribution in mind.
pos = lo
+ (((hi - lo) / (arr[hi] - arr[lo]))
* (x - arr[lo]));
// Condition of target found
if (arr[pos] == x)
return pos;
// If x is larger, x is in right sub array
if (arr[pos] < x)
return interpolationSearch(arr, pos + 1, hi,
x);
// If x is smaller, x is in left sub array
if (arr[pos] > x)
return interpolationSearch(arr, lo, pos - 1,
x);
}
return -1;
}
// Driver Code
public static void main(String[] args)
{
// Array of items on which search will
// be conducted.
int arr[] = { 10, 12, 13, 16, 18, 19, 20, 21,
22, 23, 24, 33, 35, 42, 47 };
int n = arr.length;
// Element to be searched
int x = 18;
int index = interpolationSearch(arr, 0, n - 1, x);
// If element was found
if (index != -1)
System.out.println("Element found at index "
+ index);
else
System.out.println("Element not found.");
}
}
// This code is contributed by equbalzeeshan
# Python3 program to implement
# interpolation search
# with recursion
# If x is present in arr[0..n-1], then
# returns index of it, else returns -1.
def interpolationSearch(arr, lo, hi, x):
# Since array is sorted, an element present
# in array must be in range defined by corner
if (lo <= hi and x >= arr[lo] and x <= arr[hi]):
# Probing the position with keeping
# uniform distribution in mind.
pos = lo + ((hi - lo) // (arr[hi] - arr[lo]) *
(x - arr[lo]))
# Condition of target found
if arr[pos] == x:
return pos
# If x is larger, x is in right subarray
if arr[pos] < x:
return interpolationSearch(arr, pos + 1,
hi, x)
# If x is smaller, x is in left subarray
if arr[pos] > x:
return interpolationSearch(arr, lo,
pos - 1, x)
return -1
# Driver code
# Array of items in which
# search will be conducted
arr = [10, 12, 13, 16, 18, 19, 20,
21, 22, 23, 24, 33, 35, 42, 47]
n = len(arr)
# Element to be searched
x = 18
index = interpolationSearch(arr, 0, n - 1, x)
if index != -1:
print("Element found at index", index)
else:
print("Element not found")
# This code is contributed by Hardik Jain
// C# program to implement
// interpolation search
using System;
class GFG{
// If x is present in
// arr[0..n-1], then
// returns index of it,
// else returns -1.
static int interpolationSearch(int []arr, int lo,
int hi, int x)
{
int pos;
// Since array is sorted, an element
// present in array must be in range
// defined by corner
if (lo <= hi && x >= arr[lo] &&
x <= arr[hi])
{
// Probing the position
// with keeping uniform
// distribution in mind.
pos = lo + (((hi - lo) /
(arr[hi] - arr[lo])) *
(x - arr[lo]));
// Condition of
// target found
if(arr[pos] == x)
return pos;
// If x is larger, x is in right sub array
if(arr[pos] < x)
return interpolationSearch(arr, pos + 1,
hi, x);
// If x is smaller, x is in left sub array
if(arr[pos] > x)
return interpolationSearch(arr, lo,
pos - 1, x);
}
return -1;
}
// Driver Code
public static void Main()
{
// Array of items on which search will
// be conducted.
int []arr = new int[]{ 10, 12, 13, 16, 18,
19, 20, 21, 22, 23,
24, 33, 35, 42, 47 };
// Element to be searched
int x = 18;
int n = arr.Length;
int index = interpolationSearch(arr, 0, n - 1, x);
// If element was found
if (index != -1)
Console.WriteLine("Element found at index " +
index);
else
Console.WriteLine("Element not found.");
}
}
// This code is contributed by equbalzeeshan
<script>
// Javascript program to implement Interpolation Search
// If x is present in arr[0..n-1], then returns
// index of it, else returns -1.
function interpolationSearch(arr, lo, hi, x){
let pos;
// Since array is sorted, an element present
// in array must be in range defined by corner
if (lo <= hi && x >= arr[lo] && x <= arr[hi]) {
// Probing the position with keeping
// uniform distribution in mind.
pos = lo + Math.floor(((hi - lo) / (arr[hi] - arr[lo])) * (x - arr[lo]));;
// Condition of target found
if (arr[pos] == x){
return pos;
}
// If x is larger, x is in right sub array
if (arr[pos] < x){
return interpolationSearch(arr, pos + 1, hi, x);
}
// If x is smaller, x is in left sub array
if (arr[pos] > x){
return interpolationSearch(arr, lo, pos - 1, x);
}
}
return -1;
}
// Driver Code
let arr = [10, 12, 13, 16, 18, 19, 20, 21,
22, 23, 24, 33, 35, 42, 47];
let n = arr.length;
// Element to be searched
let x = 18
let index = interpolationSearch(arr, 0, n - 1, x);
// If element was found
if (index != -1){
document.write(`Element found at index ${index}`)
}else{
document.write("Element not found");
}
// This code is contributed by _saurabh_jaiswal
</script>
<?php
// PHP program to implement $erpolation search
// with recursion
// If x is present in arr[0..n-1], then returns
// index of it, else returns -1.
function interpolationSearch($arr, $lo, $hi, $x)
{
// Since array is sorted, an element present
// in array must be in range defined by corner
if ($lo <= $hi && $x >= $arr[$lo] && $x <= $arr[$hi]) {
// Probing the position with keeping
// uniform distribution in mind.
$pos = (int)($lo
+ (((double)($hi - $lo)
/ ($arr[$hi] - $arr[$lo]))
* ($x - $arr[$lo])));
// Condition of target found
if ($arr[$pos] == $x)
return $pos;
// If x is larger, x is in right sub array
if ($arr[$pos] < $x)
return interpolationSearch($arr, $pos + 1, $hi,
$x);
// If x is smaller, x is in left sub array
if ($arr[$pos] > $x)
return interpolationSearch($arr, $lo, $pos - 1,
$x);
}
return -1;
}
// Driver Code
// Array of items on which search will
// be conducted.
$arr = array(10, 12, 13, 16, 18, 19, 20, 21, 22, 23, 24, 33,
35, 42, 47);
$n = sizeof($arr);
$x = 47; // Element to be searched
$index = interpolationSearch($arr, 0, $n - 1, $x);
// If element was found
if ($index != -1)
echo "Element found at index ".$index;
else
echo "Element not found.";
return 0;
#This code is contributed by Susobhan Akhuli
?>
OutputElement found at index 4
Time Complexity: O(log2(log2 n)) for the average case, and O(n) for the worst case
Auxiliary Space Complexity: O(1)
Another approach:-
This is the iteration approach for the interpolation search.
- Step1: In a loop, calculate the value of "pos" using the probe position formula.
- Step2: If it is a match, return the index of the item, and exit.
- Step3: If the item is less than arr[pos], calculate the probe position of the left sub-array. Otherwise, calculate the same in the right sub-array.
- Step4: Repeat until a match is found or the sub-array reduces to zero.
Below is the implementation of the algorithm.
C++
// C++ program to implement interpolation search by using iteration approach
#include<bits/stdc++.h>
using namespace std;
int interpolationSearch(int arr[], int n, int x)
{
// Find indexes of two corners
int low = 0, high = (n - 1);
// Since array is sorted, an element present
// in array must be in range defined by corner
while (low <= high && x >= arr[low] && x <= arr[high])
{
if (low == high)
{if (arr[low] == x) return low;
return -1;
}
// Probing the position with keeping
// uniform distribution in mind.
int pos = low + (((double)(high - low) /
(arr[high] - arr[low])) * (x - arr[low]));
// Condition of target found
if (arr[pos] == x)
return pos;
// If x is larger, x is in upper part
if (arr[pos] < x)
low = pos + 1;
// If x is smaller, x is in the lower part
else
high = pos - 1;
}
return -1;
}
// Main function
int main()
{
// Array of items on whighch search will
// be conducted.
int arr[] = {10, 12, 13, 16, 18, 19, 20, 21,
22, 23, 24, 33, 35, 42, 47};
int n = sizeof(arr)/sizeof(arr[0]);
int x = 18; // Element to be searched
int index = interpolationSearch(arr, n, x);
// If element was found
if (index != -1)
cout << "Element found at index " << index;
else
cout << "Element not found.";
return 0;
}
//this code contributed by Ajay Singh
// Java program to implement interpolation
// search with recursion
import java.util.*;
class GFG {
// If x is present in arr[0..n-1], then returns
// index of it, else returns -1.
public static int interpolationSearch(int arr[], int lo,
int hi, int x)
{
int pos;
if (lo <= hi && x >= arr[lo] && x <= arr[hi]) {
// Probing the position with keeping
// uniform distribution in mind.
pos = lo
+ (((hi - lo) / (arr[hi] - arr[lo]))
* (x - arr[lo]));
// Condition of target found
if (arr[pos] == x)
return pos;
// If x is larger, x is in right sub array
if (arr[pos] < x)
return interpolationSearch(arr, pos + 1, hi,
x);
// If x is smaller, x is in left sub array
if (arr[pos] > x)
return interpolationSearch(arr, lo, pos - 1,
x);
}
return -1;
}
// Driver Code
public static void main(String[] args)
{
// Array of items on which search will
// be conducted.
int arr[] = { 10, 12, 13, 16, 18, 19, 20, 21,
22, 23, 24, 33, 35, 42, 47 };
int n = arr.length;
// Element to be searched
int x = 18;
int index = interpolationSearch(arr, 0, n - 1, x);
// If element was found
if (index != -1)
System.out.println("Element found at index "
+ index);
else
System.out.println("Element not found.");
}
}
# Python equivalent of above C++ code
# Python program to implement interpolation search by using iteration approach
def interpolationSearch(arr, n, x):
# Find indexes of two corners
low = 0
high = (n - 1)
# Since array is sorted, an element present
# in array must be in range defined by corner
while low <= high and x >= arr[low] and x <= arr[high]:
if low == high:
if arr[low] == x:
return low;
return -1;
# Probing the position with keeping
# uniform distribution in mind.
pos = int(low + (((float(high - low)/( arr[high] - arr[low])) * (x - arr[low]))))
# Condition of target found
if arr[pos] == x:
return pos
# If x is larger, x is in upper part
if arr[pos] < x:
low = pos + 1;
# If x is smaller, x is in lower part
else:
high = pos - 1;
return -1
# Main function
if __name__ == "__main__":
# Array of items on whighch search will
# be conducted.
arr = [10, 12, 13, 16, 18, 19, 20, 21,
22, 23, 24, 33, 35, 42, 47]
n = len(arr)
x = 18 # Element to be searched
index = interpolationSearch(arr, n, x)
# If element was found
if index != -1:
print ("Element found at index",index)
else:
print ("Element not found")
// C# program to implement interpolation search by using
// iteration approach
using System;
class Program
{
// Interpolation Search function
static int InterpolationSearch(int[] arr, int n, int x)
{
int low = 0;
int high = n - 1;
while (low <= high && x >= arr[low] && x <= arr[high])
{
if (low == high)
{
if (arr[low] == x)
return low;
return -1;
}
int pos = low + (int)(((float)(high - low) / (arr[high] - arr[low])) * (x - arr[low]));
if (arr[pos] == x)
return pos;
if (arr[pos] < x)
low = pos + 1;
else
high = pos - 1;
}
return -1;
}
// Main function
static void Main(string[] args)
{
int[] arr = {10, 12, 13, 16, 18, 19, 20, 21, 22, 23, 24, 33, 35, 42, 47};
int n = arr.Length;
int x = 18;
int index = InterpolationSearch(arr, n, x);
if (index != -1)
Console.WriteLine("Element found at index " + index);
else
Console.WriteLine("Element not found");
}
}
// This code is contributed by Susobhan Akhuli
// JavaScript program to implement interpolation search by using iteration approach
function interpolationSearch(arr, n, x) {
// Find indexes of two corners
let low = 0;
let high = n - 1;
// Since array is sorted, an element present
// in array must be in range defined by corner
while (low <= high && x >= arr[low] && x <= arr[high]) {
if (low == high) {
if (arr[low] == x) {
return low;
}
return -1;
}
// Probing the position with keeping
// uniform distribution in mind.
let pos = Math.floor(low + (((high - low) / (arr[high] - arr[low])) * (x - arr[low])));
// Condition of target found
if (arr[pos] == x) {
return pos;
}
// If x is larger, x is in upper part
if (arr[pos] < x) {
low = pos + 1;
}
// If x is smaller, x is in lower part
else {
high = pos - 1;
}
}
return -1;
}
// Main function
let arr = [10, 12, 13, 16, 18, 19, 20, 21, 22, 23, 24, 33, 35, 42, 47];
let n = arr.length;
let x = 18; // Element to be searched
let index = interpolationSearch(arr, n, x);
// If element was found
if (index != -1) {
console.log("Element found at index", index);
} else {
console.log("Element not found");
}
OutputElement found at index 4
Time Complexity: O(log2(log2 n)) for the average case, and O(n) for the worst case
Auxiliary Space Complexity: O(1)
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Two Sum - Pair Closest to 0Given an integer array arr[], the task is to find the maximum sum of two elements such that sum is closest to zero. Note: In case if we have two of more ways to form sum of two elements closest to zero return the maximum sum.Examples:Input: arr[] = [-8, 5, 2, -6]Output: -1Explanation: The min absolu
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Pair with the given differenceGiven an unsorted array and an integer x, the task is to find if there exists a pair of elements in the array whose absolute difference is x. Examples: Input: arr[] = [5, 20, 3, 2, 50, 80], x = 78Output: YesExplanation: The pair is {2, 80}.Input: arr[] = [90, 70, 20, 80, 50], x = 45Output: NoExplana
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Kth smallest element in a row-wise and column-wise sorted 2D arrayGiven an n x n matrix, every row and column is sorted in non-decreasing order. Given a number K where K lies in the range [1, n*n], find the Kth smallest element in the given 2D matrix.Example:Input: mat =[[10, 20, 30, 40], [15, 25, 35, 45], [24, 29, 37, 48], [32, 33, 39, 50]]K = 3Output: 20Explanat
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Find common elements in three sorted arraysGiven three sorted arrays in non-decreasing order, print all common elements in non-decreasing order across these arrays. If there are no such elements return an empty array. In this case, the output will be -1.Note: In case of duplicate common elements, print only once.Examples: Input: arr1[] = [1,
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Ceiling in a sorted arrayGiven a sorted array and a value x, find index of the ceiling of x. The ceiling of x is the smallest element in an array greater than or equal to x. Note: In case of multiple occurrences of ceiling of x, return the index of the first occurrence.Examples : Input: arr[] = [1, 2, 8, 10, 10, 12, 19], x
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Floor in a Sorted ArrayGiven a sorted array and a value x, find the element of the floor of x. The floor of x is the largest element in the array smaller than or equal to x.Examples:Input: arr[] = [1, 2, 8, 10, 10, 12, 19], x = 5Output: 1Explanation: Largest number less than or equal to 5 is 2, whose index is 1Input: arr[
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Bitonic Point - Maximum in Increasing Decreasing ArrayGiven an array arr[] of integers which is initially strictly increasing and then strictly decreasing, the task is to find the bitonic point, that is the maximum value in the array. Note: Bitonic Point is a point in bitonic sequence before which elements are strictly increasing and after which elemen
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Given Array of size n and a number k, find all elements that appear more than n/k timesGiven an array of size n and an integer k, find all elements in the array that appear more than n/k times. Examples:Input: arr[ ] = [3, 4, 2, 2, 1, 2, 3, 3], k = 4Output: [2, 3]Explanation: Here n/k is 8/4 = 2, therefore 2 appears 3 times in the array that is greater than 2 and 3 appears 3 times in
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Medium problems on Searching algorithms
3 Sum - Find All Triplets with Zero SumGiven an array arr[], the task is to find all possible indices {i, j, k} of triplet {arr[i], arr[j], arr[k]} such that their sum is equal to zero and all indices in a triplet should be distinct (i != j, j != k, k != i). We need to return indices of a triplet in sorted order, i.e., i < j < k.Ex
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Find the element before which all the elements are smaller than it, and after which all are greaterGiven an array, find an element before which all elements are equal or smaller than it, and after which all the elements are equal or greater.Note: Print -1, if no such element exists.Examples:Input: arr[] = [5, 1, 4, 3, 6, 8, 10, 7, 9]Output: 6 Explanation: 6 is present at index 4. All elements on
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Largest pair sum in an arrayGiven an unsorted of distinct integers, find the largest pair sum in it. For example, the largest pair sum is 74. If there are less than 2 elements, then we need to return -1.Input : arr[] = {12, 34, 10, 6, 40}, Output : 74Input : arr[] = {10, 10, 10}, Output : 20Input arr[] = {10}, Output : -1[Naiv
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K’th Smallest Element in Unsorted ArrayGiven an array arr[] of N distinct elements and a number K, where K is smaller than the size of the array. Find the K'th smallest element in the given array. Examples:Input: arr[] = {7, 10, 4, 3, 20, 15}, K = 3 Output: 7Input: arr[] = {7, 10, 4, 3, 20, 15}, K = 4 Output: 10 Table of Content[Naive Ap
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Search in a Sorted and Rotated ArrayGiven a sorted and rotated array arr[] of n distinct elements, the task is to find the index of given key in the array. If the key is not present in the array, return -1. Examples: Input: arr[] = [5, 6, 7, 8, 9, 10, 1, 2, 3], key = 3Output: 8Explanation: 3 is present at index 8 in arr[].Input: arr[]
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Minimum in a Sorted and Rotated ArrayGiven a sorted array of distinct elements arr[] of size n that is rotated at some unknown point, the task is to find the minimum element in it. Examples: Input: arr[] = [5, 6, 1, 2, 3, 4]Output: 1Explanation: 1 is the minimum element present in the array.Input: arr[] = [3, 1, 2]Output: 1Explanation:
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Find a Fixed Point (Value equal to index) in a given arrayGiven an array of n distinct integers sorted in ascending order, the task is to find the First Fixed Point in the array. Fixed Point in an array is an index i such that arr[i] equals i. Note that integers in the array can be negative. Note: If no Fixed Point is present in the array, print -1.Example
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K Mmost Frequent Words in a FileGiven a book of words and an integer K. Assume you have enough main memory to accommodate all words. Design a dynamic data structure to find the top K most frequent words in a book. The structure should allow new words to be added in main memory.Examples:Input: fileData = "Welcome to the world of Ge
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Closest K Elements in a Sorted ArrayYou are given a sorted array arr[] containing unique integers, a number k, and a target value x. Your goal is to return exactly k elements from the array that are closest to x, excluding x itself if it is present in the array.An element a is closer to x than b if:|a - x| < |b - x|, or|a - x| == |
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2 Sum - Pair Sum Closest to Target using Binary SearchGiven an array arr[] of n integers and an integer target, the task is to find a pair in arr[] such that it’s sum is closest to target.Note: Return the pair in sorted order and if there are multiple such pairs return the pair with maximum absolute difference. If no such pair exists return an empty ar
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Find the closest pair from two sorted arraysGiven two arrays arr1[0...m-1] and arr2[0..n-1], and a number x, the task is to find the pair arr1[i] + arr2[j] such that absolute value of (arr1[i] + arr2[j] - x) is minimum. Example: Input: arr1[] = {1, 4, 5, 7}; arr2[] = {10, 20, 30, 40}; x = 32Output: 1 and 30Input: arr1[] = {1, 4, 5, 7}; arr2[]
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Find three closest elements from given three sorted arraysGiven three sorted arrays A[], B[] and C[], find 3 elements i, j and k from A, B and C respectively such that max(abs(A[i] - B[j]), abs(B[j] - C[k]), abs(C[k] - A[i])) is minimized. Here abs() indicates absolute value. Example : Input : A[] = {1, 4, 10} B[] = {2, 15, 20} C[] = {10, 12} Output: 10 15
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Search in an Array of Rational Numbers without floating point arithmeticGiven a sorted array of rational numbers, where each rational number is represented in the form p/q (where p is the numerator and q is the denominator), the task is to find the index of a given rational number x in the array. If the number does not exist in the array, return -1.Examples: Input: arr[
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Hard problems on Searching algorithms
Median of two sorted arrays of same sizeGiven 2 sorted arrays a[] and b[], each of size n, the task is to find the median of the array obtained after merging a[] and b[]. Note: Since the size of the merged array will always be even, the median will be the average of the middle two numbers.Input: a[] = [1, 12, 15, 26, 38], b[] = [2, 13, 17
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Search in an almost sorted arrayGiven a sorted integer array arr[] consisting of distinct elements, where some elements of the array are moved to either of the adjacent positions, i.e. arr[i] may be present at arr[i-1] or arr[i+1].Given an integer target. You have to return the index ( 0-based ) of the target in the array. If targ
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Find position of an element in a sorted array of infinite numbersGiven a sorted array arr[] of infinite numbers. The task is to search for an element k in the array.Examples:Input: arr[] = [3, 5, 7, 9, 10, 90, 100, 130, 140, 160, 170], k = 10Output: 4Explanation: 10 is at index 4 in array.Input: arr[] = [2, 5, 7, 9], k = 3Output: -1Explanation: 3 is not present i
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Pair Sum in a Sorted and Rotated ArrayGiven an array arr[] of size n, which is sorted and then rotated around an unknown pivot, the task is to check whether there exists a pair of elements in the array whose sum is equal to a given target value.Examples : Input: arr[] = [11, 15, 6, 8, 9, 10], target = 16Output: trueExplanation: There is
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K’th Smallest/Largest Element in Unsorted Array | Worst case Linear TimeGiven an array of distinct integers arr[] and an integer k. The task is to find the k-th smallest element in the array. For better understanding, k refers to the element that would appear in the k-th position if the array were sorted in ascending order. Note: k will always be less than the size of t
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K'th largest element in a streamGiven an input stream of n integers, represented as an array arr[], and an integer k. After each insertion of an element into the stream, you need to determine the kth largest element so far (considering all elements including duplicates). If k elements have not yet been inserted, return -1 for that
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Best First Search (Informed Search)Best First Search is a heuristic search algorithm that selects the most promising node for expansion based on an evaluation function. It prioritizes nodes in the search space using a heuristic to estimate their potential. By iteratively choosing the most promising node, it aims to efficiently naviga
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