Minimum count of increment of K size subarrays required to form a given Array

Last Updated : 15 Jul, 2025

Given an array arr[] and an integer K, the task is to find the minimum number of operations required to change an array B of size N containing all zeros such that every element of B is greater than or equal to arr. i.e., arr[i] >= B[i]. In any operation, you can choose a subarray of B of size K and increment all the elements of the subarray by 1.

Examples:

Input: arr[] = {1, 2, 3, 4, 5}, K = 2
Output: 9
Explanation:
At first B[] = {0, 0, 0, 0, 0} operations = 0
Increment subarray a[1:2] by 1 => B = {1, 1, 0, 0, 0}, operations = 1
Increment subarray a[2:3] by 1 => B = {1, 2, 1, 0, 0}, operations = 2
Increment subarray a[3:4] by 2 => B = {1, 2, 3, 2, 0}, operations = 4
Increment subarray a[4:5] by 5 => B = {1, 2, 3, 7, 5}, operations = 9
Therefore, count of such operations required is 9.

Input: arr[] = {2, 3, 1}, K = 3
Output: 3
Explanation:
Incrementing the entire array by 3

Approach: The idea is to increment the subarray of size K whenever there is a B[i] is less than arr[i] and also increment the count of such operations by 1 at each step. To increment the subarray of size K use the Difference array for Range Query update in O(1).

Below is the implementation of the above approach:

C++

// C++ implementation to find the 
// minimum number of operations 
// required to change an array of
// all zeros such that every element
// is greater than the given array
#include <bits/stdc++.h>
using namespace std;

// Function to find the minimum 
// number of operations required
// to change all the array of zeros
// such that every element is greater
// than the given array
int find_minimum_operations(int n, int b[],
                            int k)
{
    
    // Declaring the difference
    // array of size N
    int d[n + 1] = {0};

    // Number of operations
    int operations = 0, need;

    for(int i = 0; i < n; i++)
    {
        
        // First update the D[i] value 
        // with the previous value
        if (i > 0)
        {
            d[i] += d[i - 1];
        }
        
        // The index i has to be incremented
        if (b[i] > d[i])
        {
            
            // We have to perform 
            // (b[i]-d[i]) operations more
            operations += b[i] - d[i];

            need = b[i] - d[i];

            // Increment the range
            // i to i + k by need
            d[i] += need;

            // Check if i + k is valid index
            if(i + k <= n)
            {
                d[i + k]-= need;
            }
        }
    }
    cout << operations << endl;
}

// Driver Code
int main()
{
    int n = 5;
    int b[] = { 1, 2, 3, 4, 5 };
    int k = 2;
    
    // Function Call
    find_minimum_operations(n, b, k);
    
    return 0;
}

// This code is contributed by shubhamsingh10

Java

// Java implementation to find the 
// minimum number of operations 
// required to change an array of
// all zeros such that every element
// is greater than the given array
class GFG{

// Function to find the minimum 
// number of operations required
// to change all the array of zeros
// such that every element is greater
// than the given array
static void find_minimum_operations(int n, int b[],
                                    int k)
{
    
    // Declaring the difference
    // array of size N
    int d[] = new int[n + 1];

    // Number of operations
    int i, operations = 0, need;

    for(i = 0; i < n; i++)
    {
        
        // First update the D[i] value 
        // with the previous value
        if (i > 0)
        {
            d[i] += d[i - 1];
        }
        
        // The index i has to be incremented
        if (b[i] > d[i])
        {
            
            // We have to perform 
            // (b[i]-d[i]) operations more
            operations += b[i] - d[i];

            need = b[i] - d[i];

            // Increment the range
            // i to i + k by need
            d[i] += need;

            // Check if i + k is valid index
            if(i + k <= n)
            {
                d[i + k]-= need;
            }
        }
    }
    System.out.println(operations);
}

// Driver Code
public static void main (String []args)
{
    int n = 5;
    int b[] = { 1, 2, 3, 4, 5 };
    int k = 2;
    
    // Function Call
    find_minimum_operations(n, b, k);
}
}

// This code is contributed by chitranayal

Python3

# Python3 implementation to find the 
# minimum number of operations required 
# to change an array of all zeros 
# such that every element is greater than
# the given array

# Function to find the minimum 
# number of operations required
# to change all the array of zeros
# such that every element is greater
# than the given array
def find_minimum_operations(n, b, k):

    # Declaring the difference
    # array of size N
    d =[0 for i in range(n + 1)] 

    # Number of operations
    operations = 0

    for i in range(n):

        # First update the D[i] value with
        # the previous value
        d[i]+= d[i-1]

        # The index i has to be incremented
        if b[i]>d[i]:

            # We have to perform 
            # (b[i]-d[i]) operations more
            operations+=(b[i]-d[i])

            need =(b[i]-d[i]) 

            # Increment the range
            # i to i + k by need
            d[i]+= need

            # Check if i + k is valid index
            if i + k<= n:
                d[i + k]-= need

    return operations

# Driver Code
if __name__ == "__main__":
    n = 5
    b =[1, 2, 3, 4, 5]
    k = 2
    
    # Function Call
    print(find_minimum_operations(n, b, k))

// C# implementation to find the 
// minimum number of operations 
// required to change an array of
// all zeros such that every element
// is greater than the given array
using System;
class GFG{
 
// Function to find the minimum 
// number of operations required
// to change all the array of zeros
// such that every element is greater
// than the given array
static void find_minimum_operations(int n, int[] b,
                                    int k)
{
     
    // Declaring the difference
    // array of size N
    int[] d = new int[n + 1];
 
    // Number of operations
    int i, operations = 0, need;
 
    for(i = 0; i < n; i++)
    {
         
        // First update the D[i] value 
        // with the previous value
        if (i > 0)
        {
            d[i] += d[i - 1];
        }
         
        // The index i has to be incremented
        if (b[i] > d[i])
        {
             
            // We have to perform 
            // (b[i]-d[i]) operations more
            operations += b[i] - d[i];
 
            need = b[i] - d[i];
 
            // Increment the range
            // i to i + k by need
            d[i] += need;
 
            // Check if i + k is valid index
            if(i + k <= n)
            {
                d[i + k]-= need;
            }
        }
    }
    Console.Write(operations);
}
 
// Driver Code
public static void Main (string []args)
{
    int n = 5;
    int[] b = { 1, 2, 3, 4, 5 };
    int k = 2;
     
    // Function Call
    find_minimum_operations(n, b, k);
}
}
 
// This code is contributed by rock_cool

JavaScript

<script>

// Javascript implementation to find the
// minimum number of operations
// required to change an array of
// all zeros such that every element
// is greater than the given array

// Function to find the minimum
// number of operations required
// to change all the array of zeros
// such that every element is greater
// than the given array
function find_minimum_operations(n, b, k)
{

    // Declaring the difference
    // array of size N
    let d = new Array(n + 1);
    d.fill(0);

    // Number of operations
    let i, operations = 0, need;

    for(i = 0; i < n; i++)
    {

        // First update the D[i] value
        // with the previous value
        if (i > 0)
        {
            d[i] += d[i - 1];
        }

        // The index i has to be incremented
        if (b[i] > d[i])
        {

            // We have to perform
            // (b[i]-d[i]) operations more
            operations += b[i] - d[i];

            need = b[i] - d[i];

            // Increment the range
            // i to i + k by need
            d[i] += need;

            // Check if i + k is valid index
            if (i + k <= n)
            {
                d[i + k]-= need;
            }
        }
    }
    document.write(operations);
}

// Driver code
let n = 5;
let b = [ 1, 2, 3, 4, 5 ];
let k = 2;
  
// Function Call
find_minimum_operations(n, b, k);

// This code is contributed by divyesh072019

</script>

Output:

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

Analysis of Algorithms

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Minimum count of increment of K size subarrays required to form a given Array

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