Minimum cost to reach end of 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 indexes 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++

// 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;
}

Java

// 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

Python3

# 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

C#

// 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

PHP

<?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.
?>

Javascript

<script>

// JavaScript 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
            var ans = Number.MAX_VALUE;

            // Visit all possible reachable index
            for (var 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

        var a = [ 10, 30, 40, 50, 20];
        var k = 3;
        var n = a.length;
        var dp = Array(n).fill(-1);

        document.write(FindMinimumCost(0, a, n, k, dp));

// This code contributed by aashish1995

</script>

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 relaxation 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++

// 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;
}

Java

// Java implementation of the approach
import java.util.*;
class GFG
{

// Function for returning the
// min of two elements
static int min(int a, int b)
{
    return (a > b) ? b : a;
}

static int minCostJumpsDP(int []A, int k)
{
    // for calculating the number of elements
    int size = A.length;

    // Allocating Memo table and
    // initializing with INT_MAX
    int []x = new int[size];
    Arrays.fill(x, Integer.MAX_VALUE);

    // 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 < size; j++)
        {
            // Relaxing the element
            x[j] = min(x[j], x[i] +
              Math.abs(A[i] - A[j]));
        }
    }

    // return the last element in the array
    return x[size - 1];
}

// Driver Code
public static void main(String []args)
{
    int []input = { 83, 26, 37, 35, 33, 35, 56 };
    System.out.println(minCostJumpsDP(input, 3));
}
}

// This code is contributed by Rajput-Ji

Python3

# 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

C#

// C# implementation of the approach
using System;

class GFG
{

// Function for returning the
// min of two elements
static int min(int a, int b)
{
    return (a > b) ? b : a;
}

static int minCostJumpsDP(int []A, int k)
{
    // for calculating the number of elements
    int size = A.Length;

    // Allocating Memo table and
    // initializing with INT_MAX
    int []x = new int[size];
    for (int i = 0; i < size; i++)
        x[i] = int.MaxValue;
    // 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 < size; j++)
        {
            // Relaxing the element
            x[j] = min(x[j], x[i] +
            Math.Abs(A[i] - A[j]));
        }
    }

    // return the last element in the array
    return x[size - 1];
}

// Driver Code
public static void Main(String []args)
{
    int []input = { 83, 26, 37, 35, 33, 35, 56 };
    Console.WriteLine(minCostJumpsDP(input, 3));
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

    // JavaScript implementation of the approach

    // Function for returning the
    // min of two elements
    function min(a, b)
    {
        return (a > b) ? b : a;
    }

    function minCostJumpsDP(A, k)
    {
        // for calculating the number of elements
        let size = A.length;

        // Allocating Memo table and
        // initializing with INT_MAX
        let x = new Array(size);
        for (let i = 0; i < size; i++)
            x[i] = Number.MAX_VALUE;
        // Base case
        x[0] = 0;

        // For every element relax every reachable
        // element ie relax next k elements
        for (let i = 0; i < size; i++)
        {
            // reaching next k element
            for (let j = i + 1;
                     j < i + k + 1 &&
                     j < size; j++)
            {
                // Relaxing the element
                x[j] = min(x[j], x[i] +
                Math.abs(A[i] - A[j]));
            }
        }

        // return the last element in the array
        return x[size - 1];
    }

    let input = [ 83, 26, 37, 35, 33, 35, 56 ];
    document.write(minCostJumpsDP(input, 3));

</script>

Output:

69

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