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Minimize cost to reach the end of given Binary Array using jump of K length after performing given operations

  • Difficulty Level : Hard
  • Last Updated : 11 Nov, 2021

Given a binary array arr[] of N integers and an integer P, the task is to find the minimum cost to reach the end of the array from Pth index using jumps of length K. A jump to index i is valid if arr[i] = 1. The array can be modified using the following operations:

  • Replace an index having a value 0 to 1. The cost of this operation is X.
  • Delete the integer at index P. The cost of this operation is Y.

Example:

Input: arr[] = {0, 0, 0, 1, 1, 0, 0, 1, 1, 1}, P = 6, K = 2, X = 1, Y = 2
Output: 1
Explanation: In 1st operation, replace the value at index 6 to 1. Hence, arr[] = {0, 0, 0, 1, 1, 1, 0, 1, 1, 1}. Initially, the current index = P = 6. Jump from P to P + K => from 6 => 8. Again jump to the next possible index i.e, 8 => 10, which is the end of the array.

Input: arr[] = {0, 1, 0, 0, 0, 1, 0}, P = 4, K = 1, X = 2, Y = 1
Output: 4

 

Approach: The given problem can be solved based on the following observations:

  • For a given P, check if indices P, P+K, P+2K… have their value as 1. If not, then replace them with 1 and maintain their count. Hence, the final cost = count * X.
  • Upon applying the second operation i times, the starting index will become P+i. Hence the final cost = (i * Y) + (count * X).

Hence, the given problem can be solved by iterating over all possible values of i in the range [1, N-P) and calculating the cost at each step. It can be done by maintaining an array zeroes[], where zeroes[i] stores the frequency of 0’s at indices i, i+K, i+2K…

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the minimum cost to
// reach the end of the given array
int minimumCost(int arr[], int N,
                int P, int K, int X,
                int Y)
{
    // Convert P to 0-based indexing
    P = P - 1;
 
    // Vector to store the count of zeros
    // till the current index
    vector<int> zeros(N, 0);
 
    // Traverse the array and store the
    // count of zeros in vector
    for (int i = 0; i < N; i++) {
 
        // If element is 0
        if (arr[i] == 0) {
            zeros[i]++;
        }
    }
    // Iterate in the range [N-K-1, 0]
    // and increment zeros[i] by zeros[i+K]
    for (int i = N - K - 1; i >= 0; i--) {
        zeros[i] += zeros[i + K];
    }
 
    // Variable to store the min cost
    int cost = INT_MAX;
 
    // Loop to calculate the cost for all
    // values of i in range [P, N]
    for (int i = P; i < N; i++) {
        cost = min(cost,
                   ((i - P) * Y)
                       + (zeros[i] * X));
    }
 
    // Return Answer
    return cost;
}
 
// Driver Code
int main()
{
    int arr[] = { 0, 1, 0, 0, 0, 1, 0 };
    int P = 4;
    int K = 1;
    int X = 2;
    int Y = 1;
    int N = sizeof(arr) / sizeof(arr[0]);
 
    cout << minimumCost(arr, N, P, K, X, Y);
 
    return 0;
}

Java




// Java program for the above approach
 
public class GFG {
     
 
// Function to find the minimum cost to
// reach the end of the given array
static int minimumCost(int arr[], int N,
                int P, int K, int X,
                int Y)
{
    // Convert P to 0-based indexing
    P = P - 1;
 
    // Vector to store the count of zeros
    // till the current index
    int zeros[] = new int[N] ;
 
    // Traverse the array and store the
    // count of zeros in vector
    for (int i = 0; i < N; i++) {
 
        // If element is 0
        if (arr[i] == 0) {
            zeros[i]++;
        }
    }
    // Iterate in the range [N-K-1, 0]
    // and increment zeros[i] by zeros[i+K]
    for (int i = N - K - 1; i >= 0; i--) {
        zeros[i] += zeros[i + K];
    }
 
    // Variable to store the min cost
    int cost = Integer.MAX_VALUE;
 
    // Loop to calculate the cost for all
    // values of i in range [P, N]
    for (int i = P; i < N; i++) {
        cost = Math.min(cost,
                   ((i - P) * Y)
                       + (zeros[i] * X));
    }
 
    // Return Answer
    return cost;
}
 
    // Driver Code
    public static void main (String[] args) {
         
    int arr[] = { 0, 1, 0, 0, 0, 1, 0 };
    int P = 4;
    int K = 1;
    int X = 2;
    int Y = 1;
    int N = arr.length;
 
    System.out.println(minimumCost(arr, N, P, K, X, Y));
    }
}
 
// This code is contributed by AnkThon

Python3




# Python3 program for the above approach
import sys
 
# Function to find the minimum cost to
# reach the end of the given array
def minimumCost(arr, N, P, K,  X, Y) :
 
    # Convert P to 0-based indexing
    P = P - 1;
 
    # Vector to store the count of zeros
    # till the current index
    zeros = [0] * N ;
 
    # Traverse the array and store the
    # count of zeros in vector
    for i in range(N) :
 
        # If element is 0
        if (arr[i] == 0) :
            zeros[i] += 1;
 
    # Iterate in the range [N-K-1, 0]
    # and increment zeros[i] by zeros[i+K]
    for i in range( N - K - 1, -1, -1) :
        zeros[i] += zeros[i + K];
 
    # Variable to store the min cost
    cost = sys.maxsize;
 
    # Loop to calculate the cost for all
    # values of i in range [P, N]
    for i in range(P, N) :
        cost = min(cost,((i - P) * Y) + (zeros[i] * X));
 
    # Return Answer
    return cost;
 
# Driver Code
if __name__ == "__main__" :
 
    arr = [ 0, 1, 0, 0, 0, 1, 0 ];
    P = 4;
    K = 1;
    X = 2;
    Y = 1;
    N = len(arr);
 
    print(minimumCost(arr, N, P, K, X, Y));
 
   # This code is contributed by AnkThon

C#




// C# program for the above approach
using System;
 
public class GFG
{
 
    // Function to find the minimum cost to
    // reach the end of the given array
    static int minimumCost(int[] arr, int N, int P, int K, int X, int Y)
    {
       
        // Convert P to 0-based indexing
        P = P - 1;
 
        // Vector to store the count of zeros
        // till the current index
        int[] zeros = new int[N];
 
        // Traverse the array and store the
        // count of zeros in vector
        for (int i = 0; i < N; i++)
        {
 
            // If element is 0
            if (arr[i] == 0)
            {
                zeros[i]++;
            }
        }
        // Iterate in the range [N-K-1, 0]
        // and increment zeros[i] by zeros[i+K]
        for (int i = N - K - 1; i >= 0; i--)
        {
            zeros[i] += zeros[i + K];
        }
 
        // Variable to store the min cost
        int cost = int.MaxValue;
 
        // Loop to calculate the cost for all
        // values of i in range [P, N]
        for (int i = P; i < N; i++)
        {
            cost = Math.Min(cost,
                       ((i - P) * Y)
                           + (zeros[i] * X));
        }
 
        // Return Answer
        return cost;
    }
 
    // Driver Code
    public static void Main()
    {
 
        int[] arr = { 0, 1, 0, 0, 0, 1, 0 };
        int P = 4;
        int K = 1;
        int X = 2;
        int Y = 1;
        int N = arr.Length;
 
        Console.WriteLine(minimumCost(arr, N, P, K, X, Y));
    }
}
 
// This code is contributed by _Saurabh_Jaiswal

Javascript




<script>
     
// Javascript program for the above approach
 
// Function to find the minimum cost to
// reach the end of the given array
function minimumCost(arr, N, P, K, X, Y)
{
    // Convert P to 0-based indexing
    P = P - 1;
 
    // Vector to store the count of zeros
    // till the current index
    var zeros = Array(N).fill(0);
 
    // Traverse the array and store the
    // count of zeros in vector
    for (var i = 0; i < N; i++) {
 
        // If element is 0
        if (arr[i] == 0) {
            zeros[i]++;
        }
    }
    // Iterate in the range [N-K-1, 0]
    // and increment zeros[i] by zeros[i+K]
    for (var i = N - K - 1; i >= 0; i--) {
        zeros[i] += zeros[i + K];
    }
 
    // Variable to store the min cost
    var cost = 1000000000;
 
    // Loop to calculate the cost for all
    // values of i in range [P, N]
    for (var i = P; i < N; i++) {
        cost = Math.min(cost,
                   ((i - P) * Y)
                       + (zeros[i] * X));
    }
 
    // Return Answer
    return cost;
}
 
// Driver Code
var arr = [ 0, 1, 0, 0, 0, 1, 0 ];
var P = 4;
var K = 1;
var X = 2;
var Y = 1;
var N = arr.length;
document.write(minimumCost(arr, N, P, K, X, Y));
 
// This code is contributed by rutvik_56.
</script>

 
 

Output
4

 

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

 


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