Minimum operations to make sum of neighbouring elements <= X
Last Updated :
18 Jan, 2023
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++
// 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 currently 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;
}
// 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 currently 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
# 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 currently 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
// 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 currently 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 */
<script>
// Javascript implementation of the approach
// Function to return the minimum
// number of operations required
function MinOperations(n, x, arr)
{
// To store total operations required
let total = 0;
for (let i = 0; i < n; ++i)
{
// First make all elements equal to x
// which are currently greater
if (arr[i] > x)
{
let difference = arr[i] - x;
total = total + difference;
arr[i] = x;
}
}
// Left scan the array
for (let i = 1; i < n; ++i)
{
let LeftNeigbouringSum = arr[i] + arr[i - 1];
// Update the current element such that
// neighbouring sum is < x
if (LeftNeigbouringSum > x)
{
let current_diff = LeftNeigbouringSum - x;
arr[i] = Math.max(0, arr[i] - current_diff);
total = total + current_diff;
}
}
return total;
}
// Driver code
let X = 1;
let arr = [ 1, 6, 1, 2, 0, 4 ];
let N = arr.length;
document.write(MinOperations(N, X, arr)+"<br>");
// This code is contributed by rag2127
</script>
Time complexity: O(N)
Auxiliary Space: O(1)
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