Given an array arr[] of size N and an integer X. The task is to find the smallest element greater than X which is not present in the array.
Examples:
Input: arr[] = {1, 5, 10, 4, 7}, X = 4
Output: 6
6 is the smallest number greater than 4
which is not present in the array.Input: arr[] = {1, 5, 10, 6, 11}, X = 10
Output: 12
Approach:
An efficient solution is based on binary search. First sort the array. Take low as zero and high as N – 1. And search for the element X + 1. If the element at mid ( which is (low+high)/2 ) is less than or equals to the searching element then make low as mid + 1. Otherwise, make high as mid – 1. If the element at mid gets equal to the searching element then increment the searching element by one and make high as N – 1.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the smallest element greater// than x which is not present in a[]int Next_greater(int a[], int n, int x){ // Sort the array sort(a, a + n); int low = 0, high = n - 1, ans = x + 1; // Continue until low is less // than or equals to high while (low <= high) { // Find mid int mid = (low + high) / 2; // If element at mid is less than // or equals to searching element if (a[mid] <= ans) { // If mid is equals // to searching element if (a[mid] == ans) { // Increment searching element ans++; // Make high as N - 1 high = n - 1; } // Make low as mid + 1 low = mid + 1; } // Make high as mid - 1 else high = mid - 1; } // Return the next greater element return ans;}// Driver codeint main(){ int a[] = { 1, 5, 10, 4, 7 }, x = 4; int n = sizeof(a) / sizeof(a[0]); cout << Next_greater(a, n, x); return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG {// Function to return the smallest element greater// than x which is not present in a[]static int Next_greater(int a[], int n, int x){ // Sort the array Arrays.sort(a); int low = 0, high = n - 1, ans = x + 1; // Continue until low is less // than or equals to high while (low <= high) { // Find mid int mid = (low + high) / 2; // If element at mid is less than // or equals to searching element if (a[mid] <= ans) { // If mid is equals // to searching element if (a[mid] == ans) { // Increment searching element ans++; // Make high as N - 1 high = n - 1; } // Make low as mid + 1 low = mid + 1; } // Make high as mid - 1 else high = mid - 1; } // Return the next greater element return ans;}// Driver codepublic static void main(String[] args) { int a[] = { 1, 5, 10, 4, 7 }, x = 4; int n = a.length; System.out.println(Next_greater(a, n, x));}}// This code is contributed by PrinciRaj1992 |
Python3
# Python3 implementation of the approach# Function to return the smallest element # greater than x which is not present in a[]def Next_greater(a, n, x): # Sort the array a = sorted(a) low, high, ans = 0, n - 1, x + 1 # Continue until low is less # than or equals to high while (low <= high): # Find mid mid = (low + high) // 2 # If element at mid is less than # or equals to searching element if (a[mid] <= ans): # If mid is equals # to searching element if (a[mid] == ans): # Increment searching element ans += 1 # Make high as N - 1 high = n - 1 # Make low as mid + 1 low = mid + 1 # Make high as mid - 1 else: high = mid - 1 # Return the next greater element return ans# Driver codea = [1, 5, 10, 4, 7]x = 4n = len(a)print(Next_greater(a, n, x))# This code is contributed # by Mohit Kumar |
C#
// C# implementation of the approachusing System; class GFG {// Function to return the smallest element greater// than x which is not present in a[]static int Next_greater(int []a, int n, int x){ // Sort the array Array.Sort(a); int low = 0, high = n - 1, ans = x + 1; // Continue until low is less // than or equals to high while (low <= high) { // Find mid int mid = (low + high) / 2; // If element at mid is less than // or equals to searching element if (a[mid] <= ans) { // If mid is equals // to searching element if (a[mid] == ans) { // Increment searching element ans++; // Make high as N - 1 high = n - 1; } // Make low as mid + 1 low = mid + 1; } // Make high as mid - 1 else high = mid - 1; } // Return the next greater element return ans;}// Driver codepublic static void Main(String[] args) { int []a = { 1, 5, 10, 4, 7 }; int x = 4; int n = a.Length; Console.WriteLine(Next_greater(a, n, x));}} // This code is contributed by Princi Singh |
PHP
<?php// PHP implementation of the approach// Function to return the smallest element greater// than x which is not present in a[]function Next_greater($a, $n, $x){ // Sort the array sort($a); $low = 0; $high = $n - 1; $ans = $x + 1; // Continue until low is less // than or equals to high while ($low <= $high) { // Find mid $mid = ($low + $high) / 2; // If element at mid is less than // or equals to searching element if ($a[$mid] <= $ans) { // If mid is equals // to searching element if ($a[$mid] == $ans) { // Increment searching element $ans++; // Make high as N - 1 $high = $n - 1; } // Make low as mid + 1 $low = $mid + 1; } // Make high as mid - 1 else $high = $mid - 1; } // Return the next greater element return $ans;}// Driver code$a = array( 1, 5, 10, 4, 7 );$x = 4;$n = count($a);echo Next_greater($a, $n, $x);// This code is contributed by Naman_garg. ?> |
Javascript
<script>// Js implementation of the approach// Function to return the smallest element greater// than x which is not present in a[]function Next_greater( a, n, x){ // Sort the array a.sort(function(aa, bb){return aa - bb}); let low = 0, high = n - 1, ans = x + 1; // Continue until low is less // than or equals to high while (low <= high) { // Find mid let mid = Math.floor((low + high) / 2); // If element at mid is less than // or equals to searching element if (a[mid] <= ans) { // If mid is equals // to searching element if (a[mid] == ans) { // Increment searching element ans++; // Make high as N - 1 high = n - 1; } // Make low as mid + 1 low = mid + 1; } // Make high as mid - 1 else high = mid - 1; } // Return the next greater element return ans;}// Driver codelet a = [ 1, 5, 10, 4, 7 ]let x = 4;let n = a.length;document.write( Next_greater(a, n, x));</script> |
Output:
6
Time complexity: O( n*log2n ) as sorting is required for implementation.
Auxiliary Space: O(1)
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