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++ 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 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
# 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# 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
<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>
<?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.
?>
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++ 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 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 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# 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
<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)