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Minimum decrements required such that sum of all adjacent pairs in an Array does not exceed K
• Last Updated : 04 May, 2021

Given an array a[] consisting of N positive integers, and an integer K, the task is to find the minimum number of operations required to make the sum of adjacent elements less than or equal to K, where, one operation involves decreasing any array element by 1. For every ith element in the given array, the operations can be performed 0 to a[i] times. Since the answer can be large, compute it modulo 109 + 7.

Examples:

Input: a[] = {11, 3, 13, 10, 8, 17, 22}, K = 14
Output: 34
Explanation:
Minimum number of operations required to obtain the desired arrangement is as follows:

• Reduce a by 2
• Reduce a by 7
• Reduce a by 11
• Reduce a by 14

The given array is modified to the following arrangement = {11, 3, 11, 3, 8, 6, 8}
Total number of operations is 2 + 5 + 7 + 11 + 18 = 34.

Input: a[] = {1000000000, 1000000000, 1000000000, 1000000000}, K = 0
Output3: 999999979
Explanation:
Since the sum of adjacent pairs is required to be 0, all elements in the array need to be reduced to 0.
Therefore, the answer is sum of array % (109 + 7).
Sum of array is 4000000000
Therefore, the required answer is 4000000000 % 109 + 7 = 999999979

Approach:

Follow the steps below to solve the problem:

1. Iterate over the array a[] and for each adjacent pair, check if their sum is less than or equal to K. If found to be true, no changes are required.
2. For pairs with sum greater than K, follow the steps below:
• If the first element of pair exceeds K, make the value of the first element in the pair equal to K. Increase the number of operations required by value of first element – K and update the value of the first element of the pair to K.
• Now, apply the sum of pair – K operations on the second element to ensure that the sum of pair is equal to K, and update the second element of pair to K – value of first element.
3. Repeat the above steps for all the elements and print the number of operations calculated.

Below is the implementation of the above approach:

## C++

 `// C++ program to implement``// the above approach``#include``using` `namespace` `std;` `// Function to calculate the minimum``// number of operations required``int` `minimum_required_operations(``int` `arr[],``                                ``int` `n, ``int` `k)``{``    ` `    ``// Stores the total number``    ``// of operations``    ``int` `answer = 0;` `    ``long` `long` `mod = 1000000007;` `    ``// Iterate over the array``    ``for``(``int` `i = 0; i < n - 1; i++)``    ``{``        ` `        ``// If the sum of pair of adjacent``        ``// elements exceed k.``        ``if` `(arr[i] + arr[i + 1] > k)``        ``{``            ` `            ``// If current element exceeds k``            ``if` `(arr[i] > k)``            ``{``                ` `                ``// Reduce arr[i] to k``                ``answer += (arr[i] - k);``                ``arr[i] = k;``            ``}``            ` `            ``// Update arr[i + 1] accordingly``            ``answer += (arr[i] + arr[i + 1]) - k;``            ``arr[i + 1] = (k - arr[i]);` `            ``// Update answer``            ``answer %= mod;``        ``}``    ``}``    ``return` `answer;``}` `// Driver Code``int` `main()``{``    ``int` `a[] = { 9, 50, 4, 14, 42, 89 };``    ``int` `k = 10;``    ``int` `n = ``sizeof``(a) / ``sizeof``(a);``    ` `    ``cout << (minimum_required_operations(a, n, k));``    ` `    ``return` `0;``}` `// This code is contributed by chitranayal`

## Java

 `// Java program to implement``// the above approach``import` `java.util.*;` `class` `GFG{` `// Function to calculate the minimum``// number of operations required``static` `int` `minimum_required_operations(``int` `arr[],``                                       ``int` `n, ``int` `k)``{``    ` `    ``// Stores the total number``    ``// of operations``    ``int` `answer = ``0``;` `    ``long` `mod = ``1000000007``;` `    ``// Iterate over the array``    ``for``(``int` `i = ``0``; i < n - ``1``; i++)``    ``{``        ` `        ``// If the sum of pair of adjacent``        ``// elements exceed k.``        ``if` `(arr[i] + arr[i + ``1``] > k)``        ``{``            ` `            ``// If current element exceeds k``            ``if` `(arr[i] > k)``            ``{``                ` `                ``// Reduce arr[i] to k``                ``answer += (arr[i] - k);``                ``arr[i] = k;``            ``}``            ` `            ``// Update arr[i + 1] accordingly``            ``answer += (arr[i] + arr[i + ``1``]) - k;``            ``arr[i + ``1``] = (k - arr[i]);` `            ``// Update answer``            ``answer %= mod;``        ``}``    ``}``    ``return` `answer;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `a[] = { ``9``, ``50``, ``4``, ``14``, ``42``, ``89` `};``    ``int` `k = ``10``;``    ``int` `n = a.length;``    ` `    ``System.out.print(``        ``minimum_required_operations(a, n, k));``}``}` `// This code is contributed by Amit Katiyar`

## Python3

 `# Python3 Program to implement``# the above approach` `# Function to calculate the minimum``# number of operations required``def` `minimum_required_operations(arr, n, k):` `    ``# Stores the total number``    ``# of operations``    ``answer ``=` `0` `    ``mod ``=` `10` `*``*` `9` `+` `7` `    ``# Iterate over the array``    ``for` `i ``in` `range``(n ``-` `1``):` `        ``# If the sum of pair of adjacent``        ``# elements exceed k.``        ``if` `arr[i] ``+` `arr[i ``+` `1``] > k:` `            ``# If current element exceeds k``            ``if` `arr[i] > k:``              ` `              ``# Reduce arr[i] to k``                ``answer ``+``=` `(arr[i] ``-` `k)``                ``arr[i] ``=` `k` `            ``# Update arr[i + 1] accordingly``            ``answer ``+``=` `(arr[i] ``+` `arr[i ``+` `1``]) ``-` `k``            ``arr[i ``+` `1``] ``=` `(k ``-` `arr[i])` `            ``# Update answer``            ``answer ``%``=` `mod``            ` `    ``return` `answer`  `# Driver Code` `a ``=` `[``9``, ``50``, ``4``, ``14``, ``42``, ``89``]``k ``=` `10` `print``(minimum_required_operations(a, ``len``(a), k))`

## C#

 `// C# program to implement``// the above approach``using` `System;``class` `GFG{` `  ``// Function to calculate the minimum``  ``// number of operations required``  ``static` `int` `minimum_required_operations(``int``[] arr, ``int` `n,``                                         ``int` `k)``  ``{` `    ``// Stores the total number``    ``// of operations``    ``int` `answer = 0;` `    ``long` `mod = 1000000007;` `    ``// Iterate over the array``    ``for` `(``int` `i = 0; i < n - 1; i++)``    ``{` `      ``// If the sum of pair of adjacent``      ``// elements exceed k.``      ``if` `(arr[i] + arr[i + 1] > k)``      ``{` `        ``// If current element exceeds k``        ``if` `(arr[i] > k)``        ``{` `          ``// Reduce arr[i] to k``          ``answer += (arr[i] - k);``          ``arr[i] = k;``        ``}` `        ``// Update arr[i + 1] accordingly``        ``answer += (arr[i] + arr[i + 1]) - k;``        ``arr[i + 1] = (k - arr[i]);` `        ``// Update answer``        ``answer = (``int``)(answer % mod);``      ``}``    ``}``    ``return` `answer;``  ``}` `  ``// Driver Code``  ``public` `static` `void` `Main(String[] args)``  ``{``    ``int``[] a = { 9, 50, 4, 14, 42, 89 };``    ``int` `k = 10;``    ``int` `n = a.Length;` `    ``Console.Write(minimum_required_operations(a, n, k));``  ``}``}` `// This code is contributed by gauravrajput1`

## Javascript

 ``
Output:
`178`

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

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