Given an array arr[] consisting of N positive integers, the task is to find the minimum cost required to either cross the array or reach the end of the array by only moving to indices (i + 2) and (i – 1) from the ith index.
Examples:
Input: arr[] = {5, 1, 2, 10, 100}
Output: 18
Explanation:
Optimal cost path (0 based indexing): 0 ? 2 ? 1 ? 3 ? 5
Therefore, the minimum cost = 5 + 2 + 1 + 10 = 18.Input: arr[] = {9, 4, 6, 8, 5}
Output: 20
Explanation:
Optimal cost path (0 based indexing): 0 ? 2 ? 4
Therefore, the minimum cost = 9 + 6 + 5 = 20
Naive Approach: The given problem can be solved based on the following observations:
- Since all costs are positive, it will never be an optimal option to move more than one step backward, hence to reach a particular index i of the array, either jump directly from the (i – 2)th index or jump from (i – 1)th to (i + 1)th index, i.e. (2 jumps forward), followed by 1 backward jump, i.e. from (i + 1)th index to ith index.
- Now, traverse from the end of the array recursively and for the elements at indices (i – 2) and (i – 1), calculate the minimum cost of the two. Therefore, the minimum cost to cross the array can be calculated using the following recurrence relation:
minCost(index) = minimum(minCost(index – 2) + arr[i], minCost(index – 1) + arr[i] + arr[i + 1])
Time Complexity: O(2N)
Auxiliary Space: O(1)
Efficient Approach: The approach discussed above has both Optimal Substructure and Overlapping Subproblems. Therefore it can be optimized by either using Memoization or Tabulation. Follow the steps below to solve the problem:
- Initialize an array dp[], where dp[i] stores the minimum cost to reach the ith index.
- Initialize dp[0] = arr[0] as the cost to reach the 0th index, which is equal to the value at the 0th index itself. Update dp[1] = arr[0] + arr[1] + arr[2], as to reach the 1st index, jump from the 0th index to 2nd index indexn to the 1st index.
- Iterate over the range [2, N – 2] using a variable i and update dp[i] as the minimum of (dp[i – 2] + arr[i]) and (dp[i – 1] + arr[i] + arr[i + 1]).
- For the last index (N – 1), update dp[N – 1] as minimum of (dp[N – 3] + arr[N – 1]) and (dp[N – 2]).
- After completing the above steps, print the value of dp[N – 1] as the result.
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 an arrayvoid minCost(int arr[], int n){ // Base Case: When N < 3 if (n < 3) { cout << arr[0]; return; } // Store the results in table int* dp = new int[n]; // Initialize base cases dp[0] = arr[0]; dp[1] = dp[0] + arr[1] + arr[2]; // Iterate over the range[2, N - 2] // to construct the dp array for (int i = 2; i < n - 1; i++) dp[i] = min(dp[i - 2] + arr[i], dp[i - 1] + arr[i] + arr[i + 1]); // Handle case for the last index, i.e. N - 1 dp[n - 1] = min(dp[n - 2], dp[n - 3] + arr[n - 1]); // Print the answer cout << dp[n - 1];}// Driver Codeint main(){ int arr[] = { 9, 4, 6, 8, 5 }; int N = sizeof(arr) / sizeof(arr[0]); minCost(arr, N); return 0;} |
Java
// Java Program to implement// the above approachimport java.io.*;import java.util.*;class GFG{ // Function to find the minimum cost // to reach the end of an array static void minCost(int arr[], int n) { // Base Case: When N < 3 if (n < 3) { System.out.println(arr[0]); return; } // Store the results in table int dp[] = new int[n]; // Initialize base cases dp[0] = arr[0]; dp[1] = dp[0] + arr[1] + arr[2]; // Iterate over the range[2, N - 2] // to construct the dp array for (int i = 2; i < n - 1; i++) dp[i] = Math.min(dp[i - 2] + arr[i], dp[i - 1] + arr[i] + arr[i + 1]); // Handle case for the last index, i.e. N - 1 dp[n - 1] = Math.min(dp[n - 2], dp[n - 3] + arr[n - 1]); // Print the answer System.out.println(dp[n - 1]); } // Driver Code public static void main(String[] args) { int arr[] = { 9, 4, 6, 8, 5 }; int N = arr.length; minCost(arr, N); }}// This code is contributed by Kingash. |
Python3
# Python 3 program for the above approach# Function to find the minimum cost# to reach the end of an arraydef minCost(arr, n): # Base Case: When N < 3 if (n < 3): print(arr[0]) return # Store the results in table dp = [0] * n # Initialize base cases dp[0] = arr[0] dp[1] = dp[0] + arr[1] + arr[2] # Iterate over the range[2, N - 2] # to construct the dp array for i in range(2, n - 1): dp[i] = min(dp[i - 2] + arr[i], dp[i - 1] + arr[i] + arr[i + 1]) # Handle case for the last index, i.e. N - 1 dp[n - 1] = min(dp[n - 2], dp[n - 3] + arr[n - 1]) # Print the answer print(dp[n - 1])# Driver Codeif __name__ == "__main__": arr = [9, 4, 6, 8, 5] N = len(arr) minCost(arr, N) # This code is contributed by ukasp. |
C#
// C# Program to implement// the above approachusing System;public class GFG{ // Function to find the minimum cost // to reach the end of an array static void minCost(int []arr, int n) { // Base Case: When N < 3 if (n < 3) { Console.WriteLine(arr[0]); return; } // Store the results in table int []dp = new int[n]; // Initialize base cases dp[0] = arr[0]; dp[1] = dp[0] + arr[1] + arr[2]; // Iterate over the range[2, N - 2] // to construct the dp array for (int i = 2; i < n - 1; i++) dp[i] = Math.Min(dp[i - 2] + arr[i], dp[i - 1] + arr[i] + arr[i + 1]); // Handle case for the last index, i.e. N - 1 dp[n - 1] = Math.Min(dp[n - 2], dp[n - 3] + arr[n - 1]); // Print the answer Console.WriteLine(dp[n - 1]); } // Driver Code public static void Main(string[] args) { int []arr = { 9, 4, 6, 8, 5 }; int N = arr.Length; minCost(arr, N); }}// This code is contributed by AnkThon |
Javascript
<script>// Javascript program for the above approach // Function to find the minimum cost // to reach the end of an array function minCost(arr, n) { // Base Case: When N < 3 if (n < 3) { document.write(arr[0]); return; } // Store the results in table let dp = []; // Initialize base cases dp[0] = arr[0]; dp[1] = dp[0] + arr[1] + arr[2]; // Iterate over the range[2, N - 2] // to construct the dp array for (let i = 2; i < n - 1; i++) dp[i] = Math.min(dp[i - 2] + arr[i], dp[i - 1] + arr[i] + arr[i + 1]); // Handle case for the last index, i.e. N - 1 dp[n - 1] = Math.min(dp[n - 2], dp[n - 3] + arr[n - 1]); // Print the answer document.write(dp[n - 1]); }// Driver Code let arr = [ 9, 4, 6, 8, 5 ]; let N = arr.length; minCost(arr, N);</script> |
Output
20
Time Complexity: O(N)
Auxiliary Space: O(N)
Algorithm:
- Initialize two variables prevPrev and prev as arr[0] and arr[0]+arr[1]+arr[2] respectively.
- Iterate over the range [2, N-1] and for each index i:
a. Calculate the minimum cost to reach this index by comparing dp[i-1]+arr[i]+arr[i+1] and dp[i-2]+arr[i].
b. Update prevPrev as prev and prev as dp[i] for the next iteration. - The minimum cost to reach the end of the array is the minimum of prevPrev and prev.
C++
#include <bits/stdc++.h>using namespace std;// Function to find the minimum cost// to reach the end of an arrayvoid minCost(int arr[], int n){ // Base Case: When N < 3 if (n < 3) { cout << arr[0]; return; } // Initialize variables prevPrev and prev int prevPrev = arr[0]; int prev = arr[0] + arr[1] + arr[2]; // Iterate over the range[2, N - 1] // to construct the dp array for (int i = 2; i < n - 1; i++) { int curr = min(prevPrev + arr[i], prev + arr[i] + arr[i + 1]); prevPrev = prev; prev = curr; } // Print the answer cout << min(prevPrev, prev);}// Driver Codeint main(){ int arr[] = { 9, 4, 6, 8, 5 }; int N = sizeof(arr) / sizeof(arr[0]); minCost(arr, N); return 0;} |
Java
import java.lang.*;class Main { public static void minCost(int[] arr, int n) { // Base Case: When N < 3 if (n < 3) { System.out.println(arr[0]); return; } // Initialize variables prevPrev and prev int prevPrev = arr[0]; int prev = arr[0] + arr[1] + arr[2]; // Iterate over the range[2, N - 1] // to construct the dp array for (int i = 2; i < n - 1; i++) { int curr = Math.min(prevPrev + arr[i], prev + arr[i] + arr[i + 1]); prevPrev = prev; prev = curr; } // Print the answer System.out.println(Math.min(prevPrev, prev)); } public static void main(String[] args) { int[] arr = { 9, 4, 6, 8, 5 }; int N = arr.length; minCost(arr, N); }} |
Python3
def minCost(arr, n): # Base Case: When N < 3 if n < 3: return arr[0] # Initialize variables prevPrev and prev prevPrev = arr[0] prev = arr[0] + arr[1] + arr[2] # Iterate over the range[2, N - 1] # to construct the dp array for i in range(2, n - 1): curr = min(prevPrev + arr[i], prev + arr[i] + arr[i + 1]) prevPrev = prev prev = curr # Print the answer return min(prevPrev, prev)arr = [9, 4, 6, 8, 5]N = len(arr)print(minCost(arr, N)) |
Javascript
// Function to find the minimum cost// to reach the end of an arrayfunction minCost(arr, n) { // Base Case: When N < 3 if (n < 3) { console.log(arr[0]); return; } // Initialize variables prevPrev and prev let prevPrev = arr[0]; let prev = arr[0] + arr[1] + arr[2]; // Iterate over the range[2, N - 1] // to construct the dp array for (let i = 2; i < n - 1; i++) { let curr = Math.min(prevPrev + arr[i], prev + arr[i] + arr[i + 1]); prevPrev = prev; prev = curr; } // Print the answer console.log(Math.min(prevPrev, prev));}// Driver Codeconst arr = [9, 4, 6, 8, 5];const N = arr.length;minCost(arr, N); |
Output
15
Complexity Analysis:
Time Complexity: O(N), where N is the size of the array. We are iterating over the array once to calculate the minimum cost to reach each index.
Space Complexity: O(1). We are not using any extra space for storing the intermediate results. We are just using two variables prevPrev and prev to store the results for the previous iterations.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
