Minimum cost to reach end of array array when a maximum jump of K index is allowed

Given an array arr[] of N integers and an integer K, one can move from an index i to any other j if j ≤ i + k. The cost of moving from one index i to the other index j is abs(arr[i] – arr[j]). Initially we start from the index 0 and we need to reach the last index i.e. N – 1. The task is to reach the last index in the minimum cost possible.

Examples:

Input: arr[] = {10, 30, 40, 50, 20}, k = 3
Output: 30
0 -> 1 -> 4
the total cost will be: |10-30| + |30-20| = 30



Input: arr[] = {40, 10, 20, 70, 80, 10}, k = 4
Output: 30

Approach 1: The problem can be solved using Dynamic Programming. We start from index 0 and we can visit any of the index from i+1 to i+k, hence the minimum cost of all the paths will be stored in dp[i]. Once we reach N-1, it will be our base case. Use memoization to memoize the states, so that we do not need to revisit the state again to reduce complexity.

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 cost
// to reach the last index
int FindMinimumCost(int ind, int a[],
                    int n, int k, int dp[])
{
  
    // If we reach the last index
    if (ind == (n - 1))
        return 0;
  
    // Already visited state
    else if (dp[ind] != -1)
        return dp[ind];
    else {
  
        // Initially maximum
        int ans = INT_MAX;
  
        // Visit all possible reachable index
        for (int i = 1; i <= k; i++) {
  
            // If inside range
            if (ind + i < n)
                ans = min(ans, abs(a[ind + i] - a[ind])
                          + FindMinimumCost(ind + i, a,
                                             n, k, dp));
  
            // We cannot move any further
            else
                break;
        }
  
        // Memoize
        return dp[ind] = ans;
    }
}
  
// Driver Code
int main()
{
    int a[] = { 10, 30, 40, 50, 20 };
    int k = 3;
    int n = sizeof(a) / sizeof(a[0]);
    int dp[n];
    memset(dp, -1, sizeof dp);
    cout << FindMinimumCost(0, a, n, k, dp);
  
    return 0;
}

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Java

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// Java implementation of the approach
import java.util.*;
  
class GfG 
  
// Function to return the minimum cost 
// to reach the last index 
static int FindMinimumCost(int ind, int a[], 
                        int n, int k, int dp[]) 
  
    // If we reach the last index 
    if (ind == (n - 1)) 
        return 0
  
    // Already visited state 
    else if (dp[ind] != -1
        return dp[ind]; 
    else
  
        // Initially maximum 
        int ans = Integer.MAX_VALUE; 
  
        // Visit all possible reachable index 
        for (int i = 1; i <= k; i++)
        
  
            // If inside range 
            if (ind + i < n) 
                ans = Math.min(ans, Math.abs(a[ind + i] - a[ind]) + 
                            FindMinimumCost(ind + i, a, n, k, dp)); 
  
            // We cannot move any further 
            else
                break
        
  
        // Memoize 
        return dp[ind] = ans; 
    
  
// Driver Code 
public static void main(String[] args) 
    int a[] = { 10, 30, 40, 50, 20 }; 
    int k = 3
    int n = a.length; 
    int dp[] = new int[n]; 
    Arrays.fill(dp, -1); 
    System.out.println(FindMinimumCost(0, a, n, k, dp)); 
}
  
// This code is contributed by Prerna Saini

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Python3

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# Python 3 implementation of the approach
import sys
  
# Function to return the minimum cost
# to reach the last index
def FindMinimumCost(ind, a, n, k, dp):
      
    # If we reach the last index
    if (ind == (n - 1)):
        return 0
  
    # Already visited state
    elif (dp[ind] != -1):
        return dp[ind]
    else:
          
        # Initially maximum
        ans = sys.maxsize
  
        # Visit all possible reachable index
        for i in range(1, k + 1):
              
            # If inside range
            if (ind + i < n):
                ans = min(ans, abs(a[ind + i] - a[ind]) + 
                      FindMinimumCost(ind + i, a, n, k, dp))
  
            # We cannot move any further
            else:
                break
  
        # Memoize
        dp[ind] = ans
        return ans
  
# Driver Code
if __name__ == '__main__':
    a = [10, 30, 40, 50, 20]
    k = 3
    n = len(a)
    dp = [-1 for i in range(n)]
    print(FindMinimumCost(0, a, n, k, dp))
  
# This code is contributed by
# Surendra_Gangwar

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C#

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// C# implementation of the above approach
using System;
  
class GfG 
  
    // Function to return the minimum cost 
    // to reach the last index 
    static int FindMinimumCost(int ind, int []a, 
                            int n, int k, int []dp) 
    
      
        // If we reach the last index 
        if (ind == (n - 1)) 
            return 0; 
      
        // Already visited state 
        else if (dp[ind] != -1) 
            return dp[ind]; 
              
        else
      
            // Initially maximum 
            int ans = int.MaxValue; 
      
            // Visit all possible reachable index 
            for (int i = 1; i <= k; i++)
            
      
                // If inside range 
                if (ind + i < n) 
                    ans = Math.Min(ans, Math.Abs(a[ind + i] - a[ind]) + 
                                FindMinimumCost(ind + i, a, n, k, dp)); 
      
                // We cannot move any further 
                else
                    break
            
      
            // Memoize 
            return dp[ind] = ans; 
        
    
      
    // Driver Code 
    public static void Main() 
    
        int []a = { 10, 30, 40, 50, 20 }; 
        int k = 3; 
        int n = a.Length; 
        int []dp = new int[n]; 
        for(int i = 0; i < n ; i++)
            dp[i] = -1 ;
      
        Console.WriteLine(FindMinimumCost(0, a, n, k, dp)); 
    }
}
  
// This code is contributed by Ryuga

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PHP

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<?php
// PHP implementation of the approach
  
// Function to return the minimum cost 
// to reach the last index 
function FindMinimumCost($ind, $a, $n, $k, $dp
  
    // If we reach the last index 
    if ($ind == ($n - 1)) 
        return 0; 
  
    // Already visited state 
    else if ($dp[$ind] != -1) 
        return $dp[$ind]; 
    else
    
  
        // Initially maximum 
        $ans = PHP_INT_MAX; 
  
        // Visit all possible reachable index 
        for ($i = 1; $i <= $k; $i++)
        
  
            // If inside range 
            if ($ind + $i < $n
                $ans = min($ans, abs($a[$ind + $i] - $a[$ind]) + 
                     FindMinimumCost($ind + $i, $a, $n, $k, $dp)); 
  
            // We cannot move any further 
            else
                break
        
  
        // Memoize 
        return $dp[$ind] = $ans
    
  
// Driver Code 
$a = array(10, 30, 40, 50, 20 ); 
$k = 3; 
$n = sizeof($a); 
$dp = array(); 
$dp = array_fill(0, $n, -1); 
echo(FindMinimumCost(0, $a, $n, $k, $dp)); 
  
// This code is contributed by Code_Mech.
?>

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Output:

30

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

Approach 2:
The second approach also requires the use of Dynamic programming. This approach is based on Bellman ford’s DP solution to the single-source shortest path. In Bellman’s ford SSSP, the main idea is to find the next vertex through minimizing on edges, we can do the same we minimize on abs values of two elements of an array to find the next index.

For solving any DP problems we first guess all the possible solutions to the subproblems and memoize them then choose the best solutions to the subproblem. we write Recurrence for the problem

Recurrence: DP(j) = min{DP(i) + abs(A[i] – A[j])} where i is in [0, N-1] and j is in [i + 1, j + k + 1], and k is number of jumps allowed. this can also be compared with relaxtion in SSSP. We are going to relax every next approachable index.

// base case
memo[0] = 0;

for (i = 0 to N-1)
for (j = i+1 to i+k+1)
memo[j] = min(memo[j], memo[i] + abs(A[i] – A[j]));

Below is the implementation of the Bottom-up above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
  
using namespace std;
  
// Function for returning the min of two elements
int min(int a, int b) {
    return (a > b) ? b : a;
}
  
int minCostJumpsDP(vector <int> & A, int k) {
    // for calculating the number of elements 
    int size = A.size();
  
    // Allocating Memo table and 
    // initializing with INT_MAX
    vector <int> x(size, INT_MAX);  
  
    // Base case
    x[0] = 0;
  
    // For every element relax every reachable 
    // element ie relax next k elements
    for (int i = 0; i < size; i++) {
        // reaching next k element
        for (int j = i + 1; j < i + k + 1; j++) {
            // Relaxing the element
            x[j] = min(x[j], x[i] + abs(A[i] - A[j]));
        }
    }
  
    // return the last element in the array
    return x[size - 1];
}
  
  
// Driver Code
int main()
{
    vector <int> input { 83, 26, 37, 35, 33, 35, 56 };
    cout << minCostJumpsDP(input, 3);
    return 0;
}

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Python3

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# Python3 implementation of the approach 
import sys
  
def minCostJumpsDP(A, k): 
      
    # for calculating the number of elements 
    size = len(A)
  
    # Allocating Memo table and 
    # initializing with INT_MAX 
    x = [sys.maxsize] * (size) 
  
    # Base case 
    x[0] = 0
  
    # For every element relax every reachable 
    # element ie relax next k elements 
    for i in range(size): 
          
        # reaching next k element
        j = i+1
        while j < i + k + 1 and j < size: 
              
            # Relaxing the element 
            x[j] = min(x[j], x[i] + abs(A[i] - A[j]))
            j += 1
          
    # return the last element in the array 
    return x[size - 1
  
# Driver Code 
if __name__ == "__main__":
  
    input_ = [83, 26, 37, 35, 33, 35, 56
    print(minCostJumpsDP(input_, 3))
      
# This code is contributed by Rituraj Jain

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Output:

69

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



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