Minimum operations to make sum of neighbouring elements <= X

Given an array arr[] of N elements and an integer X, the task is to find the minimum number of operations required to make sum of neighbouring elements less than the given number X. In a single operation, you can choose an element arr[i] and decrease its value by 1.

Examples:

Input: arr[] = {2, 2, 2}, X = 3
Output: 1
Decrement arr[1] by 1 and the array becomes {2, 1, 2}.
Now, 2 + 1 = 3 and 1 + 2 = 3

Input: arr[] = {1, 6, 1, 2, 0, 4}, X = 1
Output: 11

Approach: Suppose the elements of the array are a1, a2, …, an. Let’s assume a case when all a[i] are greater than X.
First we need to make all the a[i] equal to x. We will calculate the number of operations required for it.
Now, all elements will be of the form of x, x, x, x…, x N times. Here we can observe that the maximum neighbouring sum is equal to 2 * X.
Now, traverse the array from left to right, and for each i, if sum of two left neighbours that is a[i] + a[i – 1] > X then change a[i] to such a value that their net sum becomes equal to X.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the minimum
// number of operations required
int MinOperations(int n, int x, int* arr)
{
  
    // To store total operations required
    int total = 0;
    for (int i = 0; i < n; ++i) {
  
        // First make all elements equal to x
        // which are currenctly greater
        if (arr[i] > x) {
            int difference = arr[i] - x;
            total = total + difference;
            arr[i] = x;
        }
    }
  
    // Left scan the array
    for (int i = 1; i < n; ++i) {
        int LeftNeigbouringSum = arr[i] + arr[i - 1];
  
        // Update the current element such that
        // neighbouring sum is < x
        if (LeftNeigbouringSum > x) {
            int current_diff = LeftNeigbouringSum - x;
            arr[i] = max(0, arr[i] - current_diff);
            total = total + current_diff;
        }
    }
    return total;
}
  
// Driver code
int main()
{
    int X = 1;
    int arr[] = { 1, 6, 1, 2, 0, 4 };
    int N = sizeof(arr) / sizeof(int);
    cout << MinOperations(N, X, arr);
  
    return 0;
}

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Java

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// Java implementation of the approach
class GFG 
{
  
// Function to return the minimum
// number of operations required
static int MinOperations(int n, int x, int[] arr)
{
  
    // To store total operations required
    int total = 0;
    for (int i = 0; i < n; ++i) 
    {
  
        // First make all elements equal to x
        // which are currenctly greater
        if (arr[i] > x)
        {
            int difference = arr[i] - x;
            total = total + difference;
            arr[i] = x;
        }
    }
  
    // Left scan the array
    for (int i = 1; i < n; ++i)
    {
        int LeftNeigbouringSum = arr[i] + arr[i - 1];
  
        // Update the current element such that
        // neighbouring sum is < x
        if (LeftNeigbouringSum > x) 
        {
            int current_diff = LeftNeigbouringSum - x;
            arr[i] = Math.max(0, arr[i] - current_diff);
            total = total + current_diff;
        }
    }
    return total;
}
  
// Driver code
public static void main(String args[]) 
{
    int X = 1;
    int arr[] = { 1, 6, 1, 2, 0, 4 };
    int N = arr.length;
    System.out.println(MinOperations(N, X, arr));
}
}
  
// This code is contributed by 29AjayKumar

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Python

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# Python3 implementation of the approach
  
# Function to return the minimum
# number of operations required
def MinOperations(n, x, arr):
  
    # To store total operations required
    total = 0
    for i in range(n):
  
        # First make all elements equal to x
        # which are currenctly greater
        if (arr[i] > x):
            difference = arr[i] - x
            total = total + difference
            arr[i] = x
  
  
    # Left scan the array
    for i in range(n):
        LeftNeigbouringSum = arr[i] + arr[i - 1]
  
        # Update the current element such that
        # neighbouring sum is < x
        if (LeftNeigbouringSum > x):
            current_diff = LeftNeigbouringSum - x
            arr[i] = max(0, arr[i] - current_diff)
            total = total + current_diff
  
    return total
  
  
# Driver code
X = 1
arr=[1, 6, 1, 2, 0, 4 ]
N = len(arr)
print(MinOperations(N, X, arr))
  
# This code is contributed by mohit kumar 29

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C#

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// C# implementation of the approach 
using System;
      
class GFG 
{
  
// Function to return the minimum
// number of operations required
static int MinOperations(int n, int x, int[] arr)
{
  
    // To store total operations required
    int total = 0;
    for (int i = 0; i < n; ++i) 
    {
  
        // First make all elements equal to x
        // which are currenctly greater
        if (arr[i] > x)
        {
            int difference = arr[i] - x;
            total = total + difference;
            arr[i] = x;
        }
    }
  
    // Left scan the array
    for (int i = 1; i < n; ++i)
    {
        int LeftNeigbouringSum = arr[i] + arr[i - 1];
  
        // Update the current element such that
        // neighbouring sum is < x
        if (LeftNeigbouringSum > x) 
        {
            int current_diff = LeftNeigbouringSum - x;
            arr[i] = Math.Max(0, arr[i] - current_diff);
            total = total + current_diff;
        }
    }
    return total;
}
  
// Driver code
public static void Main(String []args) 
{
    int X = 1;
    int []arr = { 1, 6, 1, 2, 0, 4 };
    int N = arr.Length;
    Console.WriteLine(MinOperations(N, X, arr));
}
}
  
/* This code is contributed by PrinciRaj1992 */

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Output:

11


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