Given an array arr[], the task is to check whether it is possible to make all the elements of the array zero by the given operation. In a single operation, any two elements arr[i] and arr[j] can be decremented by one at the same time.
Examples:
Input: arr[] = {1, 2, 1, 2, 2}
Output: Yes
Decrement the 1st and the 2nd element, arr[] = {0, 1, 1, 2, 2}
Decrement the 2nd and the 3rd element, arr[] = {0, 0, 0, 2, 2}
Decrement the 4th and the 5th element, arr[] = {0, 0, 0, 1, 1}
Decrement the 4th and the 5th element, arr[] = {0, 0, 0, 0, 0}Input: arr[] = {1, 2, 3, 4, 5}
Output: No
Approach: The given array can be only be made zero if it holds the following conditions true:
- The sum of the elements of the array must be even.
- The largest elements must be less than or equal to ⌊sum / 2⌋ where sum is the sum of the other elements.
Below is the implementation of the above approach:
CPP
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function that returns true if all // the array elements can be made // 0 with the given operation bool checkZeroArray(int* arr, int n) { // Find the maximum element // and the sum int sum = 0, maximum = INT_MIN; for (int i = 0; i < n; i++) { sum = sum + arr[i]; maximum = max(maximum, arr[i]); } // Check the required condition if (sum % 2 == 0 && maximum <= sum / 2) return true; return false; } // Driver code int main() { int arr[] = { 1, 2, 1, 2, 2 }; int n = sizeof(arr) / sizeof(int); if (checkZeroArray(arr, n)) cout << "Yes"; else cout << "No"; return 0; } |
Java
// Java implementation of the above approach public class GFG { // Function that returns true if all // the array elements can be made // 0 with the given operation static boolean checkZeroArray(int []arr, int n) { // Find the maximum element // and the sum int sum = 0, maximum = Integer.MIN_VALUE; for (int i = 0; i < n; i++) { sum = sum + arr[i]; maximum = Math.max(maximum, arr[i]); } // Check the required condition if (sum % 2 == 0 && maximum <= sum / 2) return true; return false; } // Driver code public static void main (String[] args) { int arr[] = { 1, 2, 1, 2, 2 }; int n = arr.length; if (checkZeroArray(arr, n) == true) System.out.println("Yes"); else System.out.println("No"); } } // This code is contributed by AnkitRai01 |
Python
# Python3 implementation of the approach # Function that returns true if all # the array elements can be made # 0 with the given operation def checkZeroArray(arr,n): # Find the maximum element # and the sum sum = 0 maximum = -10**9 for i in range(n): sum = sum + arr[i] maximum = max(maximum, arr[i]) # Check the required condition if (sum % 2 == 0 and maximum <= sum // 2): return True return False # Driver code arr = [1, 2, 1, 2, 2] n = len(arr) if (checkZeroArray(arr, n)): print("Yes") else: print("No") # This code is contributed by mohit kumar 29 |
C#
// C# implementation of the above approach using System; class GFG { // Function that returns true if all // the array elements can be made // 0 with the given operation static bool checkZeroArray(int []arr, int n) { // Find the maximum element // and the sum int sum = 0, maximum = int.MinValue; for (int i = 0; i < n; i++) { sum = sum + arr[i]; maximum = Math.Max(maximum, arr[i]); } // Check the required condition if (sum % 2 == 0 && maximum <= sum / 2) return true; return false; } // Driver code public static void Main () { int []arr = { 1, 2, 1, 2, 2 }; int n = arr.Length; if (checkZeroArray(arr, n) == true) Console.WriteLine("Yes"); else Console.WriteLine("No"); } } // This code is contributed by AnkitRai01 |
Yes
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