Minimize subarray increments/decrements required to reduce all array elements to 0

Last Updated : 16 Apr, 2021

Given an array arr[], select any subarray and apply any one of the below operations on each element of the subarray:

Increment by one
Decrement by one

The task is to print the minimum number of above-mentioned increment/decrement operations required to reduce all array elements to 0.

Examples:

Input: arr[] = {1, 3, 4, 1}
Output: 4
Explanation:
Optimal steps to reduce all array elements to 0 are as follows:
Step 1: Select subarray [1, 3, 4, 1] and convert it to [0, 2, 3, 0], by decrementing each element by 1
Array modifies to {0, 2, 3, 0}
Step 2: Select subarray [2, 3] and convert it to [1, 2], by decrementing each element by 1
Array modifies to {0, 1, 2, 0}
Step 3: Select subarray [1, 2] and convert it to [0, 1], by decrementing each element by 1
Array modifies to {0, 0, 1, 0}
Step 4: Select subarray [1] convert it to [0]
Array modifies to {0, 0, 0, 0}
Therefore, minimum number of steps required is 4.

Input: arr[] = {-2, 0, -3, 1, 2}
Output: 5
Explanation:
Optimal steps to reduce all array elements to 0 are as follows:
Step 1: Select subarray [-2, 0, -3] and convert it to [-1, 0, -2], by incrementing each element by 1 except 0.
Array modifies to {-1, 0, -2, 1, 2}
Step 2: Select subarray [-1, 0, -2] and convert it to [0, 0, -1], by incrementing each element by 1 except 0.
Array modifies to {0, 0, -1, 1, 2}
Step 3: Select subarray [-1] convert it to [0]
Array modifies to {0, 0, 0, 1, 2}
Step 4: Select subarray [1, 2] and convert it to [0, 1], by decrementing each element by 1
Array modifies to {0, 0, 0, 0, 1}
Step 5: Select [1] convert it to [0]
Array modifies to {0, 0, 0, 0, 0}
Therefore, minimum number of steps required is 5

Approach: To solve the problem, traverse the array and find the minimum negative number and the maximum positive number. The sum of their absolute values is the minimum number of operations required.

Below is the implementation of the above approach:

C++

// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to count the minimum
// number of operations required
int minOperation(int arr[], int N)
{
  int minOp = INT_MIN;
  int minNeg = 0, maxPos = 0;
 
  // Traverse the array
  for (int i = 0; i < N; i++) 
  {
    // If array element
    // is negative
    if (arr[i] < 0) 
    {
      if (arr[i] < minNeg)
 
        // Update minimum negative
        minNeg = arr[i];
    }
    else
    {
      if (arr[i] > maxPos)
 
        // Update maximum positive
        maxPos = arr[i];
    }
  }
 
  // Return minOp
  return abs(minNeg) + maxPos;
}
 
// Driver Code
int main()
{
  int arr[] = {1, 3, 4, 1};
  int N = sizeof(arr) / sizeof(arr[0]);
  cout << minOperation(arr, N);
}
 
//This code is contributed by Rajput-Ji

Java

// Java program of the
// above approach
 
import java.util.*;
import java.lang.*;
 
class GFG {
 
    // Function to count the minimum
    // number of operations required
    static int minOperation(int[] arr)
    {
 
        int minOp = Integer.MIN_VALUE;
        int minNeg = 0, maxPos = 0;
 
        // Traverse the array
        for (int i = 0; i < arr.length; i++) {
 
            // If array element
            // is negative
            if (arr[i] < 0) {
                if (arr[i] < minNeg)
 
                    // Update minimum negative
                    minNeg = arr[i];
            }
            else {
                if (arr[i] > maxPos)
 
                    // Update maximum positive
                    maxPos = arr[i];
            }
        }
 
        // Return minOp
        return Math.abs(minNeg) + maxPos;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = { 1, 3, 4, 1 };
        System.out.println(minOperation(arr));
    }
}

Python3

# Python3 program of the
# above approach
 
import sys
 
# Function to count the minimum
# number of operations required
def minOperation(arr):
     
    minOp = sys.maxsize
    minNeg = 0
    maxPos = 0
     
    # Traverse the array
    for i in range(len(arr)):
       
        # If array element
        # is negative
        if(arr[i] < 0):
             
            if (arr[i] < minNeg):
                     
                # Update minimum negative
                minNeg = arr[i]
        else:
            if arr[i] > maxPos:
                 
                # Update maximum position
                maxPos = arr[i]
     
    # Return minOp
    return abs(minNeg) + maxPos
 
# Driver code  
if __name__=="__main__":    
    arr=[1, 3, 4, 1]
    print(minOperation(arr))
 
# This code is contributed by Rutvik_56

C#

// C# program of the
// above approach
using System;
class GFG{
 
// Function to count the minimum
// number of operations required
static int minOperation(int[] arr)
{
 
  int minOp = int.MinValue;
  int minNeg = 0, maxPos = 0;
 
  // Traverse the array
  for (int i = 0; i < arr.Length; i++) 
  {
    // If array element
    // is negative
    if (arr[i] < 0) 
    {
      if (arr[i] < minNeg)
 
        // Update minimum negative
        minNeg = arr[i];
    }
    else
    {
      if (arr[i] > maxPos)
 
        // Update maximum positive
        maxPos = arr[i];
    }
  }
 
  // Return minOp
  return Math.Abs(minNeg) + maxPos;
}
 
// Driver Code
public static void Main(String[] args)
{
  int []arr = {1, 3, 4, 1};
  Console.WriteLine(minOperation(arr));
}
}
 
// This code is contributed by Princi Singh

Javascript

<script>
 
// JavaScript program to implement
// the above approach
 
    // Function to count the minimum
    // number of operations required
    function minOperation(arr)
    {
  
        let minOp = Number.MIN_VALUE;
        let minNeg = 0, maxPos = 0;
  
        // Traverse the array
        for (let i = 0; i < arr.length; i++)
        {
  
            // If array element
            // is negative
            if (arr[i] < 0)
            {
                if (arr[i] < minNeg)
  
                    // Update minimum negative
                    minNeg = arr[i];
            }
            else
            {
                if (arr[i] > maxPos)
  
                    // Update maximum positive
                    maxPos = arr[i];
            }
        }
  
        // Return minOp
        return Math.abs(minNeg) + maxPos;
    }
  
// Driver code
    let arr = [ 1, 3, 4, 1 ];
    document.write(minOperation(arr));
      
     // This code is contributed by target_2.
</script>