Given an integer X and an array cost[] consisting of N integers, the task is to find the minimum cost to reach the end of the given array starting from the first element based on the following constraints:
- Cost to visit point i is cost[i] where 1 ? i ? N.
- Cost to skip point i is min(cost[i], X) where 1 ? i ? N.
- At most 2 points can be skipped in a row.
- First and last positions cannot be skipped.
Examples:
Input: N = 6, X = 4, cost[] = {6, 3, 9, 2, 1, 3}
Output: 19
Explanation:
Follow the steps below:
Step 1: Choose element at 1. Sum = 6
Step 2: Choose element at 2. Sum = 6 + 3 = 9
Step 3: Skip element at 3. Sum = 6 + 3 + 4 = 13
Step 4: Choose element at 4. Sum = 6 + 3 + 4 + 2 = 15
Step 5: Choose element at 5. Sum = 6 + 3 + 4 + 2 + 1 = 16
Step 6: Choose element at 6. Sum = 6 + 3 + 4 + 2 + 1 + 3 = 19
Hence, the minimum cost is 19.Input: N = 7, X = 4, cost[] = {6, 3, 9, 2, 1, 3, 4}
Output: 23
Explanation:
Follow the steps below:
Step 1: Choose element at 1. Sum = 6
Step 2: Choose element at 2. Sum = 6+3 = 9
Step 3: Skip element at 3. Sum = 6+3+4 = 13
Step 4: Choose element at 4. Sum = 6 + 3 + 4 + 2 = 15
Step 5: Choose element at 5. Sum = 6+3+4+2+1 = 16
Step 6: Choose element at 6. Sum = 6+3+4+2+1+3 = 19
Step 7: Choose element at 6. Sum = 6 + 3 + 4 + 2 + 1 + 3 + 4 = 23
Hence, the minimum cost is 23.
Naive Approach: The simplest approach is to generate all possible solutions by considering or skipping certain positions. There are two options for each element i.e., it can be skipped or can be chosen. Therefore, there can be at most 2N combinations. Check that in each combination, no more than 3 positions are skipped. Among those combinations, choose the one having the minimum cost and print the minimum cost.
Time Complexity: O(2N)
Auxiliary Space: O(N)
Efficient Approach: To optimize the above approach, the idea is to use Dynamic Programming and observe that if any position i is skipped, the cost is increased by cost[i] or X but if the cost is increased by cost[i] then it’s best to choose that position as choosing the position also increases the cost by cost[i]. This implies that the minimum cost to reach position i can be found by taking the minimum among the minimum cost to reach position (i – 1), X + the minimum cost to reach position (i – 2) and 2X + the minimum cost to reach position (i – 3).
Therefore, the dp transition is as follows:
dp[i] = cost[i] + min(dp[i-1], min(2*X + dp[i-2], 2*X + dp[i-3]))
where,
dp[i] stores the minimum answer to reach position i from position 0.
Follow the below steps to solve the problem:
- Initialize an array dp[] where dp[i] will store the minimum answer to reach position i from position 0.
- Traverse the given array cost[] over the range [0, N – 1] and at each position i, update dp[i] as:
cost[i] + min(dp[i-1], min(2*X + dp[i-2], 2*X + dp[i-3]))
- After the above steps, print dp[N – 1] which stores the answer to reach position (N – 1) from position 0.
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 array from// the first elementvoid minimumCost(int* cost, int n, int x){ // Store the results vector<int> dp(n + 2, 0); // Consider first index cost dp[0] = cost[0]; // Find answer for each position i for (int i = 1; i < n; i++) { // First Element if (i == 1) dp[i] = cost[i] + dp[i - 1]; // Second Element if (i == 2) dp[i] = cost[i] + min(dp[i - 1], x + dp[i - 2]); // For remaining element if (i >= 3) // Consider min cost for // skipping dp[i] = cost[i] + min(dp[i - 1], min(x + dp[i - 2], 2 * x + dp[i - 3])); } // Last index represents the // minimum total cost cout << dp[n - 1];}// Driver Codeint main(){ // Given X int X = 4; // Given array cost[] int cost[] = { 6, 3, 9, 2, 1, 3 }; int N = sizeof(cost) / sizeof(cost[0]); // Function Call minimumCost(cost, N, X); return 0;} |
Java
// Java program for the above approachimport java.io.*;import java.util.*;class GFG{// Function to find the minimum cost// to reach the end of the array from// the first elementstatic void minimumCost(int[] cost, int n, int x){ // Store the results int[] dp = new int[n + 2]; // Consider first index cost dp[0] = cost[0]; // Find answer for each position i for(int i = 1; i < n; i++) { // First Element if (i == 1) dp[i] = cost[i] + dp[i - 1]; // Second Element if (i == 2) dp[i] = cost[i] + Math.min(dp[i - 1], x + dp[i - 2]); // For remaining element if (i >= 3) // Consider min cost for // skipping dp[i] = cost[i] + Math.min(dp[i - 1], Math.min(x + dp[i - 2], 2 * x + dp[i - 3])); } // Last index represents the // minimum total cost System.out.println(dp[n - 1]);}// Driver Codepublic static void main(String[] args){ // Given X int X = 4; // Given array cost[] int[] cost = { 6, 3, 9, 2, 1, 3 }; int N = cost.length; // Function Call minimumCost(cost, N, X);}}// This code is contributed by akhilsaini |
Python3
# Python3 program for the above approach# Function to find the minimum cost# to reach the end of the array from# the first elementdef minimumCost(cost, n, x): # Store the results dp = [0] * (n + 2) # Consider first index cost dp[0] = cost[0] # Find answer for each position i for i in range(1, n): # First Element if (i == 1): dp[i] = cost[i] + dp[i - 1] # Second Element if (i == 2): dp[i] = cost[i] + min(dp[i - 1], x + dp[i - 2]) # For remaining element if (i >= 3): # Consider min cost for # skipping dp[i] = (cost[i] + min(dp[i - 1], min(x + dp[i - 2], 2 * x + dp[i - 3]))) # Last index represents the # minimum total cost print(dp[n - 1])# Driver Codeif __name__ == '__main__': # Given X X = 4 # Given array cost[] cost = [ 6, 3, 9, 2, 1, 3 ] N = len(cost) # Function Call minimumCost(cost, N, X)# This code is contributed by mohit kumar 29 |
C#
// C# program for the above approachusing System;class GFG{// Function to find the minimum cost// to reach the end of the array from// the first elementstatic void minimumCost(int[] cost, int n, int x){ // Store the results int[] dp = new int[n + 2]; // Consider first index cost dp[0] = cost[0]; // Find answer for each position i for(int i = 1; i < n; i++) { // First Element if (i == 1) dp[i] = cost[i] + dp[i - 1]; // Second Element if (i == 2) dp[i] = cost[i] + Math.Min(dp[i - 1], x + dp[i - 2]); // For remaining element if (i >= 3) // Consider min cost for // skipping dp[i] = cost[i] + Math.Min(dp[i - 1], Math.Min(x + dp[i - 2], 2 * x + dp[i - 3])); } // Last index represents the // minimum total cost Console.WriteLine(dp[n - 1]);}// Driver Codepublic static void Main(){ // Given X int X = 4; // Given array cost[] int[] cost = { 6, 3, 9, 2, 1, 3 }; int N = cost.Length; // Function Call minimumCost(cost, N, X);}}// This code is contributed by akhilsaini |
Javascript
<script>// Javascript program for the above approach// Function to find the minimum cost// to reach the end of the array from// the first elementfunction minimumCost(cost, n, x){ // Store the results let dp = []; // Consider first index cost dp[0] = cost[0]; // Find answer for each position i for(let i = 1; i < n; i++) { // First Element if (i == 1) dp[i] = cost[i] + dp[i - 1]; // Second Element if (i == 2) dp[i] = cost[i] + Math.min(dp[i - 1], x + dp[i - 2]); // For remaining element if (i >= 3) // Consider min cost for // skipping dp[i] = cost[i] + Math.min(dp[i - 1], Math.min(x + dp[i - 2], 2 * x + dp[i - 3])); } // Last index represents the // minimum total cost document.write(dp[n - 1]);}// Driver code// Given Xlet X = 4;// Given array cost[]let cost = [ 6, 3, 9, 2, 1, 3 ];let N = cost.length;// Function CallminimumCost(cost, N, X);// This code is contributed by splevel62</script> |
19
Time Complexity: O(N)
Auxiliary Space: O(N)
Efficient approach: Space Optimization O(1)
In previous approach the current value i.e. dp[i] is depend upon the previous three values i.e. dp[i-1] , dp[i-2] and dp[i-3]. So instead of using DP array we use variables to store these computation which will optimize the space complexity from O(N) to O(1).
Implementation Steps:
Implementation:
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 array from// the first elementvoid minimumCost(int* cost, int n, int x){ // initialize varibles with base cases int prev1= cost[0] , prev2=0, prev3=0 , curr=0; // iterate over subproblem for (int i = 1; i < n; i++) { // if only 2 numbers we need only prev1 if (i == 1){ curr = cost[i] + prev1; // get value of prev2 prev2 = curr; } // get current value from prev2 and prev1 if (i == 2){ curr = cost[i]+ min(prev2,x + prev1); // get value of prev3 prev3 = curr; } // get current value from prev3, prev2 and prev1 if (i >= 3){ curr = cost[i]+ min(prev3,min(x + prev2,2 * x + prev1)); // assigning values to iterate further prev1 = prev2; prev2 = prev3; prev3 = curr; } } // curr represents the // minimum total cost cout << curr;}// Driver Codeint main(){ // Given X int X = 4; // Given array cost[] int cost[] = { 6, 3, 9, 2, 1, 3 }; int N = sizeof(cost) / sizeof(cost[0]); // Function Call minimumCost(cost, N, X); return 0;}// this code is contributed by bhardwajji |
Java
// Java program for the above approachimport java.util.*;public class Main { // Function to find the minimum cost// to reach the end of the array from// the first elementstatic void minimumCost(int[] cost, int n, int x) { // initialize variables with base cases int prev1 = cost[0], prev2 = 0, prev3 = 0, curr = 0; // iterate over subproblem for (int i = 1; i < n; i++) { // if only 2 numbers we need only prev1 if (i == 1) { curr = cost[i] + prev1; // get value of prev2 prev2 = curr; } // get current value from prev2 and prev1 if (i == 2) { curr = cost[i] + Math.min(prev2, x + prev1); // get value of prev3 prev3 = curr; } // get current value from prev3, prev2 and prev1 if (i >= 3) { curr = cost[i] + Math.min(prev3, Math.min(x + prev2, 2 * x + prev1)); // assigning values to iterate further prev1 = prev2; prev2 = prev3; prev3 = curr; } } // curr represents the minimum total cost System.out.println(curr);}// Driver Codepublic static void main(String[] args) { // Given X int X = 4; // Given array cost[] int[] cost = { 6, 3, 9, 2, 1, 3 }; int N = cost.length; // Function Call minimumCost(cost, N, X);}} |
Python3
# Python program for the above approachdef minimumCost(cost, n, x): # initialize variables with base cases prev1 = cost[0] prev2 = 0 prev3 = 0 curr = 0 # iterate over subproblem for i in range(1, n): # if only 2 numbers we need only prev1 if i == 1: curr = cost[i] + prev1 # get value of prev2 prev2 = curr # get current value from prev2 and prev1 if i == 2: curr = cost[i] + min(prev2, x + prev1) # get value of prev3 prev3 = curr # get current value from prev3, prev2 and prev1 if i >= 3: curr = cost[i] + min(prev3, min(x + prev2, 2 * x + prev1)) # assigning values to iterate further prev1, prev2, prev3 = prev2, prev3, curr # curr represents the minimum total cost print(curr)# Driver Codeif __name__ == '__main__': # Given X X = 4 # Given array cost[] cost = [6, 3, 9, 2, 1, 3] N = len(cost) # Function Call minimumCost(cost, N, X) |
C#
using System;class GFG { static void MinimumCost(int[] cost, int n, int x) { int prev1 = cost[0]; int prev2 = 0; int prev3 = 0; int curr = 0; for (int i = 1; i < n; i++) { if (i == 1) { curr = cost[i] + prev1; prev2 = curr; } else if (i == 2) { curr = cost[i] + Math.Min(prev2, x + prev1); prev3 = curr; } else { curr = cost[i] + Math.Min(prev3, Math.Min(x + prev2, 2 * x + prev1)); prev1 = prev2; prev2 = prev3; prev3 = curr; } } Console.WriteLine(curr); } static void Main() { int X = 4; int[] cost = new int[] {6, 3, 9, 2, 1, 3}; int N = cost.Length; MinimumCost(cost, N, X); }} |
Javascript
// JavaScript program for the above approach// Function to find the minimum cost// to reach the end of the array from// the first elementfunction minimumCost(cost, n, x) { // Initialize variables with base cases let prev1 = cost[0], prev2 = 0, prev3 = 0, curr = 0; // Iterate over subproblems for (let i = 1; i < n; i++) { // If there are only 2 numbers, we need only prev1 if (i === 1) { curr = cost[i] + prev1; // Get the value of prev2 prev2 = curr; } // Get the current value from prev2 and prev1 if (i === 2) { curr = cost[i] + Math.min(prev2, x + prev1); // Get the value of prev3 prev3 = curr; } // Get the current value from prev3, prev2, and prev1 if (i >= 3) { curr = cost[i] + Math.min(prev3, Math.min(x + prev2, 2 * x + prev1)); // Assigning values to iterate further prev1 = prev2; prev2 = prev3; prev3 = curr; } } // curr represents the minimum total cost console.log(curr);}// Driver Code// Given Xconst X = 4;// Given array cost[]const cost = [6, 3, 9, 2, 1, 3];const N = cost.length;// Function CallminimumCost(cost, N, X); |
19
Time Complexity: O(N)
Auxiliary Space: O(1)
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