Minimize operations to convert Array elements to 0s

Given an array nums[] of length N containing non-negative integers, the task is to convert all elements from index 0 to N-2 to zero, after doing the below operations minimum number of times.

• In one operation select two indices i and j such that i < j  and all the elements between i and j has to be non-zero, then set nums[i] = nums[i] - 1 and nums[j] = nums[j] + 1.

Examples:

Input: N = 5, nums[] = [0, 2, 0, 2, 0]
Output: 5
Explanation: We need to do 5 above given operation 
to complete the task, and the operations are:

• Select j = 3, k = 4 nums[] becomes: [0, 2, 0, 1, 1]
• Select j = 1, k = 2 nums[] becomes: [0, 1, 1, 1, 1]
• Select j = 1, k = 4 nums[] becomes: [0, 0, 1, 1, 2]
• Select j = 2, k = 4 nums[] becomes: [0, 0, 0, 1, 3]
• Select j = 3, k = 4 nums[] becomes: [0, 0, 0, 0, 4]

Input: N = 3, nums[] = [2, 0, 0]
Output: 3
Explanation: We need to do 3 above given operation 
to complete the task, and the operations are:

• Select j = 0, k = 1 nums[] becomes: [1, 1, 0]
• Select j = 0, k = 2 nums[] becomes: [0, 1, 1]
• Select j = 1, k = 2 nums[] becomes: [0, 0, 2]

Approach: The problem can be solved based on the following idea:

• The possible way to convert all elements from 0 to N-2 to 0 is by reducing each element from the left by 1 and adding 1 each time to the last element, until leftmost element become zero and then moving to the next index. So the number of operations will be the sum of elements from 0 to N-2.
• But if there is a 0 present in between then first we have to convert that 0 into a positive number, so that we can perform the operation for the elements present in the left of that 0. 
• To convert a 0 to a positive element, 1 operation will take place, so our final answer will be the sum of elements from 0 to N-2 and the number of zeros present after the first positive element. 

Follow the steps given below to implement the approach:

• First traverse the array till you get the first non-zero element.
  • After that if element is non-zero then update ans += nums[i]
  • Else if the element is zero then update ans += 1.
• Traverse the given array till N-2 and calculate the required answer as mentioned above.
• Return the calculated answer.

Below is the implementation of the above approach.

// C++ Algorithm for the above approach
#include <iostream>
using namespace std;

// Function which returns
// minimum number of operations
// required to convert array
// into target array
int minOperations(int nums[], int N)
{
  
    // ans variable contains number of
    // operations processed after traversing
    // ith element
    int ans = 0;
    int i = 0;

    // Traversing to get first non-zero element
    while (i < N && nums[i] == 0)
        i++;
  
    // After getting first non-zero element we will
    // apply the given operation
    while (i < N - 1) {
        if (nums[i] == 0)
            ans++;
        else
            ans += nums[i];
        i++;
    }

    // Returning the ans variable
    return ans;
}

// Driver Code
int main()
{

    int N = 5;
    int nums[] = { 0, 2, 0, 2, 0 };
  
    // Function Call
    int ans = minOperations(nums, N);
    cout << ans;
    return 0;
}

// This code is contributed by Kdheeraj.

// Java algorithm of the above approach

import java.util.*;

class GFG {

    // Driver Code
    public static void main(String[] args)
    {
        int N = 5;
        int[] nums = { 0, 2, 0, 2, 0 };
        // Function Call
        int ans = minOperations(nums, N);
        System.out.println(ans);
    }

    // Function which returns
    // minimum number of operations
    // required to convert array
    // into target array
    public static int minOperations(int[] nums, int N)
    {
        // ans variable contains number of
        // operations processed after traversing
        // ith element
        int ans = 0;
        int i = 0;

        // Traversing to get first non-zero element
        while (i < N && nums[i] == 0)
            i++;
        // After getting first non-zero element we will
        // apply the given operation
        while (i < N - 1) {
            if (nums[i] == 0)
                ans++;
            else
                ans += nums[i];
            i++;
        }

        // Returning the ans variable
        return ans;
    }
}

# Python code to implement the above approach

# Function which returns
# minimum number of operations
# required to convert array
# into target array
def minOperations(nums, N) :
  
    # ans variable contains number of
    # operations processed after traversing
    # ith element
    ans = 0
    i = 0

    # Traversing to get first non-zero element
    while (i < N and nums[i] == 0) :
        i += 1
  
    # After getting first non-zero element we will
    # apply the given operation
    while (i < N - 1) :
        if (nums[i] == 0) :
            ans += 1
        else :
            ans += nums[i]
        i += 1
    

    # Returning the ans variable
    return ans

# Driver Code
if __name__ == "__main__":

    N = 5
    nums = [ 0, 2, 0, 2, 0 ]
  
    # Function Call
    ans = minOperations(nums, N)
    print(ans)
    
    # This code is contributed by sanjoy_62.

// C# program for above approach:
using System;
class GFG {

  // Function which returns
  // minimum number of operations
  // required to convert array
  // into target array
  public static int minOperations(int[] nums, int N)
  {
    
    // ans variable contains number of
    // operations processed after traversing
    // ith element
    int ans = 0;
    int i = 0;

    // Traversing to get first non-zero element
    while (i < N && nums[i] == 0)
      i++;
    // After getting first non-zero element we will
    // apply the given operation
    while (i < N - 1) {
      if (nums[i] == 0)
        ans++;
      else
        ans += nums[i];
      i++;
    }

    // Returning the ans variable
    return ans;
  }

  // Driver Code
  public static void Main()
  {
    int N = 5;
    int[] nums = { 0, 2, 0, 2, 0 };
    
    // Function Call
    int ans = minOperations(nums, N);
    Console.Write(ans);

  }
}

// This code is contributed by code_hunt.

<script>
// Javascript Algorithm for the above approach

// Function which returns
// minimum number of operations
// required to convert array
// into target array
function minOperations(nums, N)
{
  
    // ans variable contains number of
    // operations processed after traversing
    // ith element
    let ans = 0;
    let i = 0;

    // Traversing to get first non-zero element
    while (i < N && nums[i] == 0)
        i++;
  
    // After getting first non-zero element we will
    // apply the given operation
    while (i < N - 1) {
        if (nums[i] == 0)
            ans++;
        else
            ans += nums[i];
        i++;
    }

    // Returning the ans variable
    return ans;
}

// Driver Code
    let N = 5;
    let nums = [ 0, 2, 0, 2, 0 ];
  
    // Function Call
    let ans = minOperations(nums, N);
    document.write(ans);
    
    // This code is contributed by satwik4409.
    </script>

5

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

kdheeraj

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Article Tags :

• Greedy
• Mathematical
• DSA
• Arrays

Practice Tags :

• Arrays
• Greedy
• Mathematical