Given an array arr[] of non-negative integers of size N, the task is to find an integer H such that at least K integers in the array are greater or equal to K.
Examples:
Input: arr[] = [3, 0, 6, 1, 5]
Output: 3
Explanation: There are 3 number greater than or equal to 3 in the array i.e. 3, 6 and 5.
Input: arr[] = [9, 10, 7, 5, 0, 10, 2, 0]
Output: 5
Approach: Using Hashing
The number integer K can not be greater than the size of arr[]. So, maintain the frequency of each element in a frequency array(hash table). Then traverse the frequency array from the end and return the first index that matches the condition.
Below is the implementation of the above approach.
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;int getNumberK(vector<int>& S){ // vector to store freq. vector<int> freq(S.size() + 1, 0); // Filling freq vector. for (int i = 0; i < S.size(); i++) { if (S[i] < S.size()) freq[S[i]]++; else freq[S.size()]++; } int total = 0; // Finding K number. for (int i = S.size(); i >= 0; i--) { total += freq[i]; if (total >= i) return i; } // No K number found. return 0;}// Driver codeint main(){ vector<int> arr{ 3, 0, 6, 1, 5 }; cout << getNumberK(arr) << '\n'; return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{ static int getNumberK(ArrayList<Integer> S){ // To store freq. int[] freq = new int[S.size() + 1]; // Filling freq vector. for(int i = 0; i < S.size(); i++) { if (S.get(i) < S.size()) freq[S.get(i)]++; else freq[S.size()]++; } int total = 0; // Finding K number. for(int i = S.size(); i >= 0; i--) { total += freq[i]; if (total >= i) return i; } // No K number found. return 0;}// Driver codepublic static void main(String[] args){ ArrayList<Integer> arr = new ArrayList<>(Arrays.asList(3, 0, 6, 1, 5)); System.out.println(getNumberK(arr));}}// This code is contributed by offbeat |
Python3
# Python3 program for the above approachdef getNumberK(S): # Vector to store freq. freq = [0] * (len(S) + 1) # Filling freq vector. for i in range(len(S)): if (S[i] < len(S)): freq[S[i]] += 1 else: freq[len(S)] += 1 total = 0 # Finding K number. for i in range(len(S), -1, -1): total += freq[i] if (total >= i): return i # No K number found. return 0# Driver code arr = [ 3, 0, 6, 1, 5 ] print(getNumberK(arr))# This code is contributed by code_hunt |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{ static int getNumberK(List<int> S){ // To store freq. int[] freq = new int[S.Count + 1]; // Filling freq vector. for(int i = 0; i < S.Count; i++) { if (S[i] < S.Count) freq[S[i]]++; else freq[S.Count]++; } int total = 0; // Finding K number. for(int i = S.Count; i >= 0; i--) { total += freq[i]; if (total >= i) return i; } // No K number found. return 0;}// Driver codepublic static void Main(String[] args){ List<int> arr = new List<int>{ 3, 0, 6, 1, 5 }; Console.WriteLine(getNumberK(arr));}}// This code is contributed by Amit Katiyar |
Javascript
<script>// Javascript program for// the above approachfunction getNumberK(S){ // To store freq. let freq = Array.from({length: S.length + 1}, (_, i) => 0); // Filling freq vector. for(let i = 0; i < S.length; i++) { if (S[i] < S.length) freq[S[i]]++; else freq[S.length]++; } let total = 0; // Finding K number. for(let i = S.length; i >= 0; i--) { total += freq[i]; if (total >= i) return i; } // No K number found. return 0;}// Driver code let arr = [3, 0, 6, 1, 5]; document.write(getNumberK(arr)); // This code is contributed by sourabhghosh0416. </script> |
Output:
3
Time Complexity O(N)
Space Complexity O(N)
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