Given an array arr[] of N integers and an integer S. The task is to find an element K in the array such that if all the elements from the array > K are made equal to K then the sum of all the elements of the resultant array becomes equal to S. If it is not possible to find such an element then print -1 .
Examples:
Input: M = 15, arr[] = {12, 3, 6, 7, 8}
Output: 3
Resultant array = {3, 3, 3, 3, 3}
Sum of the array elements = 15 = SInput: M = 5, arr[] = {1, 3, 2, 5, 8}
Output: 1
Approach: Sort the array. Traverse the array considering that the value of K is equal to the current element and then check if sum of the previous elements + (K * number of remaining elements) = S. If yes then print the value of K, if no such element found then print -1 in the end.
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 required element int getElement(int a[], int n, int S) { // Sort the array sort(a, a + n); int sum = 0; for (int i = 0; i < n; i++) { // If current element // satisfies the condition if (sum + (a[i] * (n - i)) == S) return a[i]; sum += a[i]; } // No element found return -1; } // Driver code int main() { int S = 5; int a[] = { 1, 3, 2, 5, 8 }; int n = sizeof(a) / sizeof(a[0]); cout << getElement(a, n, S); return 0; } |
Java
//Java implementation of the approach import java.util.Arrays; class GFG { // Function to return the required element static int getElement(int a[], int n, int S) { // Sort the array Arrays.sort(a); int sum = 0; for (int i = 0; i < n; i++) { // If current element // satisfies the condition if (sum + (a[i] * (n - i)) == S) return a[i]; sum += a[i]; } // No element found return -1; } // Driver code public static void main(String[] args) { int S = 5; int a[] = { 1, 3, 2, 5, 8 }; int n = a.length; System.out.println(getElement(a, n, S)); } } // This code is contributed by Mukul singh. |
Python 3
# Python3 implementation of the approach # Function to return the required element def getElement(a, n, S) : # Sort the array a.sort(); sum = 0; for i in range(n) : # If current element # satisfies the condition if (sum + (a[i] * (n - i)) == S) : return a[i]; sum += a[i]; # No element found return -1; # Driver Code if __name__ == "__main__" : S = 5; a = [ 1, 3, 2, 5, 8 ]; n = len(a) ; print(getElement(a, n, S)) ; # This code is contributed by Ryuga |
C#
// C# implementation of the approach using System; class GFG { // Function to return the required element static int getElement(int[] a, int n, int S) { // Sort the array Array.Sort(a); int sum = 0; for (int i = 0; i < n; i++) { // If current element // satisfies the condition if (sum + (a[i] * (n - i)) == S) return a[i]; sum += a[i]; } // No element found return -1; } // Driver code public static void Main() { int S = 5; int[] a = { 1, 3, 2, 5, 8 }; int n = a.Length; Console.WriteLine(getElement(a, n, S)); } } // This code is contributed by Mukul singh. |
PHP
<?php // PHP implementation of the approach // Function to return the required element function getElement($a, $n, $S) { // Sort the array sort($a, 0); $sum = 0; for ($i = 0; $i < $n; $i++) { // If current element // satisfies the condition if ($sum + ($a[$i] * ($n - $i)) == $S) return $a[$i]; $sum += $a[$i]; } // No element found return -1; } // Driver code $S = 5; $a = array(1, 3, 2, 5, 8); $n = sizeof($a); echo getElement($a, $n, $S); // This code is contributed // by Akanksha Rai ?> |
1
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
