Check if all array elements can be reduced to 0 by repeatedly reducing pairs of consecutive elements by their minimum

Given an array arr[] consisting of N integers, the task is to check if it is possible to reduce the array elements to 0 by repeatedly subtracting the minimum of any pair of consecutive array elements from both the elements in the pair. If it is possible, then print “Yes”. Otherwise, print “No”.

Examples:

Input: arr[] = {2, 3, 3, 4, 2}
Output: Yes
Explanation:
One of the possible way is:
Select the pair of indices (0, 1) and perform the operation, which modifies the array as arr[] = {0, 1, 3, 4, 2}.
Select the pair of indices (1, 2), and perform the operation which modifies the array as arr[] = {0, 0, 2, 4, 2}.
Select the pair of indices (2, 3), and perform the operation which modifies the array as arr[] = {0, 0, 0, 2, 2}.
Select the pair of indices (3, 4), and perform the operation which modifies the array as arr[] = {0, 0, 0, 0, 0}.
Therefore, it is possible to convert the array elements to zero. So print “YES”

Input: arr[] = {243, 12, 11}
Output: No
Explanation:
It is impossible to convert every array elements to zero.

Approach: The given problem can be solved based on the following observations: 

• Suppose the arr[] = {a, b, c, d, e, f}, then it can be observed that a should be lesser than or equal to b i.e., a ? b to satisfy the above condition. Otherwise, the element a will remain greater than 0.
• Now modifying the array as arr[] = {0, b – a, c, d, e}, b – a is the leftmost element so the same condition as above applies to it as well i.e., b – a ? c.
• So it can be observed that after repeating the above process in the end, the sum of elements at even indices must equal to the sum of elements at odd indices.

Follow the steps below to solve the problem:

• Traverse the range [1, N-1] and check if arr[i] < arr[i – 1], then print “NO” and break. Otherwise, decrement arr[i] by arr[i – 1].
• After completing the above step, and if none of the above cases satisfies, then check if arr[N – 1] = 0 or not. If found to be true, then print “YES”. Otherwise, print “NO”.
• Traverse the given array arr[] over the range [0, N – 2] using the variable i and perform the following:
  • If the value of arr[i] is less than arr[i + 1] then print “No” as all the array elements can’t be reduced to 0.
  • Otherwise, decrement arr[i + 1] by arr[i].
• After completing the above steps, if none of the above cases satisfy and the value of arr[N – 1] is 0, then print “Yes”. Otherwise, print “No”.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to check if it is possible
// to convert the array
void checkPossible(int* arr, int n)
{
    // Traverse the array range [1, N-1]
    for (int i = 1; i < n; i++) {

        // If arr[i] < arr[i-1]
        if (arr[i] < arr[i - 1]) {
            cout << "No\n";
            return;
        }

        // Otherwise
        else {

            // Decrement arr[i] by arr[i-1]
            arr[i] -= arr[i - 1];
            arr[i - 1] = 0;
        }
    }

    // If arr[n - 1] is not equal to zero
    if (arr[n - 1] == 0) {
        cout << "Yes\n";
    }

    // Otherwise
    else {
        cout << "No\n";
    }
}

// Driver Code
int main()
{
    int arr[] = { 2, 3, 3, 4, 2 };
    int N = sizeof(arr) / sizeof(arr[0]);

    // Function Call
    checkPossible(arr, N);

    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
class GFG
{

// Function to check if it is possible
// to convert the array
static void checkPossible(int[] arr, int n)
{

    // Traverse the array range [1, N-1]
    for (int i = 1; i < n; i++)
    {

        // If arr[i] < arr[i-1]
        if (arr[i] < arr[i - 1])
        {
            System.out.print("No\n");
            return;
        }

        // Otherwise
        else
        {

            // Decrement arr[i] by arr[i-1]
            arr[i] -= arr[i - 1];
            arr[i - 1] = 0;
        }
    }

    // If arr[n - 1] is not equal to zero
    if (arr[n - 1] == 0)
    {
        System.out.print("Yes\n");
    }

    // Otherwise
    else
    {
        System.out.print("No\n");
    }
}

// Driver Code
public static void main(String args[])
{
    int arr[] = { 2, 3, 3, 4, 2 };
    int N = arr.length;

    // Function Call
    checkPossible(arr, N);
}
}

// This code is contributed by splevel62.

Python3

# Python program to implement
# the above approach

# Function to check if it is possible
# to convert the array
def checkPossible(arr, n):

    # Traverse the array range [1, N-1]
    for i in range(1, n):

        # If arr[i] < arr[i-1]
        if (arr[i] < arr[i - 1]):
            print("No");
            return;

        # Otherwise
        else:

            # Decrement arr[i] by arr[i-1]
            arr[i] -= arr[i - 1];
            arr[i - 1] = 0;

    # If arr[n - 1] is not equal to zero
    if (arr[n - 1] == 0):
        print("Yes");

    # Otherwise
    else:
        print("No");

# Driver Code
if __name__ == '__main__':
    arr = [2, 3, 3, 4, 2];
    N = len(arr);

    # Function Call
    checkPossible(arr, N);

# This code is contributed by shikhasingrajput

C#

// C# program to implement
// the above approach
using System;
class GFG
{

// Function to check if it is possible
// to convert the array
static void checkPossible(int[] arr, int n)
{

    // Traverse the array range [1, N-1]
    for (int i = 1; i < n; i++)
    {

        // If arr[i] < arr[i-1]
        if (arr[i] < arr[i - 1])
        {
            Console.Write("No\n");
            return;
        }

        // Otherwise
        else
        {

            // Decrement arr[i] by arr[i-1]
            arr[i] -= arr[i - 1];
            arr[i - 1] = 0;
        }
    }

    // If arr[n - 1] is not equal to zero
    if (arr[n - 1] == 0)
    {
        Console.Write("Yes\n");
    }

    // Otherwise
    else
    {
        Console.Write("No\n");
    }
}

// Driver Code
public static void Main()
{
    int[] arr = { 2, 3, 3, 4, 2 };
    int N = arr.Length;

    // Function Call
    checkPossible(arr, N);
}
}

// This code is contributed by susmitakundugoaldanga

Javascript

<script>

// JavaScript program to implement
// the above approach

    // Function to check if it is possible
    // to convert the array
    function checkPossible(arr , n) {

        // Traverse the array range [1, N-1]
        for (i = 1; i < n; i++) {

            // If arr[i] < arr[i-1]
            if (arr[i] < arr[i - 1]) {
                document.write("No\n");
                return;
            }

            // Otherwise
            else {

                // Decrement arr[i] by arr[i-1]
                arr[i] -= arr[i - 1];
                arr[i - 1] = 0;
            }
        }

        // If arr[n - 1] is not equal to zero
        if (arr[n - 1] == 0) {
            document.write("Yes\n");
        }

        // Otherwise
        else {
            document.write("No\n");
        }
    }

    // Driver Code

        var arr = [ 2, 3, 3, 4, 2 ];
        var N = arr.length;

        // Function Call
        checkPossible(arr, N);

// This code contributed by gauravrajput1

</script>

Output

Yes

Time Complexity: O(N)
Auxiliary Space: O(1)

Using brute force in python:

Approach:

• Select the pair of indices (0, 1) and perform the operation, which modifies the array as arr = [0, 1, 3, 4, 2].
• Select the pair of indices (1, 2), and perform the operation which modifies the array as arr = [0, 0, 2, 4, 2].
• Select the pair of indices (2, 3), and perform the operation which modifies the array as arr = [0, 0, 0, 2, 2].
• Select the pair of indices (3, 4), and perform the operation which modifies the array as arr = [0, 0, 0, 0, 0].

C++

#include <iostream>
#include <vector>

using namespace std;  // Add this line

string canReduceToZero(vector<int> &arr) {
    int n = arr.size();

    // Loop to find and reduce pairs of elements to zero
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            // Find the minimum value between arr[i] and arr[j]
            int temp = min(arr[i], arr[j]);

            // Subtract the minimum value from both elements
            arr[i] -= temp;
            arr[j] -= temp;
        }
    }

    // Check if all elements are reduced to zero
    for (int i = 0; i < n; i++) {
        if (arr[i] != 0) {
            return "No";
        }
    }

    return "Yes";
}

int main() {
    vector<int> arr1 = {2, 3, 3, 4, 2};
    cout << canReduceToZero(arr1) << endl; // Output: Yes

    vector<int> arr2 = {243, 12, 11};
    cout << canReduceToZero(arr2) << endl; // Output: No

    return 0;
}

Java

import java.util.Arrays;

public class Main {
    public static void main(String[] args) {
        int[] arr1 = {2, 3, 3, 4, 2};
        System.out.println(canReduceToZero(arr1)); // Output: Yes

        int[] arr2 = {243, 12, 11};
        System.out.println(canReduceToZero(arr2)); // Output: No
    }

    public static String canReduceToZero(int[] arr) {
        int n = arr.length;

        // Loop to find and reduce pairs of elements to zero
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                // Find the minimum value between arr[i] and arr[j]
                int temp = Math.min(arr[i], arr[j]);

                // Subtract the minimum value from both elements
                arr[i] -= temp;
                arr[j] -= temp;
            }
        }

        // Check if all elements are reduced to zero
        if (Arrays.stream(arr).anyMatch(val -> val != 0)) {
            return "No";
        }

        return "Yes";
    }
}

Python3

def can_reduce_to_zero(arr):
    n = len(arr)
    for i in range(n):
        for j in range(i+1, n):
            temp = min(arr[i], arr[j])
            arr[i] -= temp
            arr[j] -= temp
    for i in range(n):
        if arr[i] != 0:
            return "No"
    return "Yes"

arr = [2, 3, 3, 4, 2]
print(can_reduce_to_zero(arr)) # Output: Yes

arr = [243, 12, 11]
print(can_reduce_to_zero(arr)) # Output: No

C#

using System;
using System.Collections.Generic;

class Program
{
    static string CanReduceToZero(List<int> arr)
    {
        int n = arr.Count;

        // Loop to find and reduce pairs of elements to zero
        for (int i = 0; i < n; i++)
        {
            for (int j = i + 1; j < n; j++)
            {
                // Find the minimum value between arr[i] and arr[j]
                int temp = Math.Min(arr[i], arr[j]);

                // Subtract the minimum value from both elements
                arr[i] -= temp;
                arr[j] -= temp;
            }
        }

        // Check if all elements are reduced to zero
        foreach (int element in arr)
        {
            if (element != 0)
            {
                return "No";
            }
        }

        return "Yes";
    }

    static void Main()
    {
        List<int> arr1 = new List<int> { 2, 3, 3, 4, 2 };
        Console.WriteLine(CanReduceToZero(arr1)); // Output: Yes

        List<int> arr2 = new List<int> { 243, 12, 11 };
        Console.WriteLine(CanReduceToZero(arr2)); // Output: No
    }
}

Javascript

// Function to check if it's possible to reduce elements to zero
function canReduceToZero(arr) {
  const n = arr.length;

  // Loop to find and reduce pairs of elements to zero
  for (let i = 0; i < n; i++) {
    for (let j = i + 1; j < n; j++) {
      // Find the minimum value between arr[i] and arr[j]
      const temp = Math.min(arr[i], arr[j]);

      // Subtract the minimum value from both elements
      arr[i] -= temp;
      arr[j] -= temp;
    }
  }

  // Check if all elements are reduced to zero
  for (let i = 0; i < n; i++) {
    if (arr[i] !== 0) {
      return 'No';
    }
  }

  return 'Yes';
}

// Test cases
const arr1 = [2, 3, 3, 4, 2];
console.log(canReduceToZero([...arr1])); // Output: Yes

const arr2 = [243, 12, 11];
console.log(canReduceToZero([...arr2])); // Output: No

Output

Yes
No

The time complexity of the can_reduce_to_zero function is O(n^2), where n is the length of the input array arr. This is because the function uses two nested loops to perform the pairwise operations between the elements of the array.

The space complexity of the function is O(1), because it only uses a constant amount of additional memory to store the loop variables and the temporary variable temp. The input array arr is modified in-place, without creating any additional arrays.