Given an array arr[] and an integer N, the task is to find a number K from the given array such that if all the elements in the arr greater than K are changed to K then the sum of all the elements in the resulting array will be N. Print -1 if it is not possible.
Examples:
Input: arr[] = {3, 1, 10, 8, 4}, N = 16
Output: 4
If all the elements greater than 4 are changed to 4 then the resulting array will be {3, 1, 4, 4, 4}
Hence the sum will be 16 which is the required value of N.Input: arr[] = {3, 1, 10, 8, 4}, N = 11
Output: -1
Approach: The idea is to use sorting to solve the above problem.
- First, sort the array in ascending order.
- Then traverse the array and for each element arr[i].
- Assume that all the elements after it has been changed to arr[i] and calculate the sum.
- If it is equal to N then return arr[i].
- If the end of the array is reached then print -1.
Below is the implementation of the above approach:
CPP
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to return K such that changing // all elements greater than K to K will // make array sum N otherwise return -1 int findK(int arr[], int size, int N) { // Sorting the array in increasing order sort(arr, arr + size); int temp_sum = 0; // Loop through all the elements of the array for (int i = 0; i < size; i++) { temp_sum += arr[i]; // Checking if sum of array equals N if (N - temp_sum == arr[i] * (size - i - 1)) { return arr[i]; } } return -1; } // Driver code int main() { int arr[] = { 3, 1, 10, 4, 8 }; int size = sizeof(arr) / sizeof(int); int N = 16; cout << findK(arr, size, N); return 0; } |
Java
// Java implementation of the approach import java.util.*; class GFG { // Function to return K such that changing // all elements greater than K to K will // make array sum N otherwise return -1 static int findK(int arr[], int size, int N) { // Sorting the array in increasing order Arrays.sort(arr); int temp_sum = 0; // Loop through all the elements of the array for (int i = 0; i < size; i++) { temp_sum += arr[i]; // Checking if sum of array equals N if (N - temp_sum == arr[i] * (size - i - 1)) { return arr[i]; } } return -1; } // Driver code public static void main (String[] args) { int []arr = { 3, 1, 10, 4, 8 }; int size = arr.length; int N = 16; System.out.print(findK(arr, size, N)); } } // This code is contributed by AnkitRai01 |
Python
# Python3 implementation of the approach # Function to return K such that changing # all elements greater than K to K will # make array sum N otherwise return -1 def findK(arr, size, N): # Sorting the array in increasing order arr = sorted(arr) temp_sum = 0 # Loop through all the elements of the array for i in range(size): temp_sum += arr[i] # Checking if sum of array equals N if (N - temp_sum == arr[i] * (size - i - 1)): return arr[i] return -1 # Driver code arr = [3, 1, 10, 4, 8] size = len(arr) N = 16 print(findK(arr, size, N)) # This code is contributed by mohit kumar 29 |
C#
// C# implementation of the above approach using System; class GFG { // Function to return K such that changing // all elements greater than K to K will // make array sum N otherwise return -1 static int findK(int []arr, int size, int N) { // Sorting the array in increasing order Array.Sort(arr); int temp_sum = 0; // Loop through all the elements of the array for (int i = 0; i < size; i++) { temp_sum += arr[i]; // Checking if sum of array equals N if (N - temp_sum == arr[i] * (size - i - 1)) { return arr[i]; } } return -1; } // Driver code public static void Main() { int []arr = { 3, 1, 10, 4, 8 }; int size = arr.Length; int N = 16; Console.Write(findK(arr, size, N)); } } // This code is contributed by AnkitRai01 |
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