Minimum cost to reach the top of the floor by climbing stairs

19 June 2025

2

Given N non-negative integers which signifies the cost of the moving from each stair. Paying the cost at i-th step, you can either climb one or two steps. Given that one can start from the 0-the step or 1-the step, the task is to find the minimum cost to reach the top of the floor(N+1) by climbing N stairs.

Examples:

Input: a[] = { 16, 19, 10, 12, 18 }
Output: 31
Start from 19 and then move to 12.

Input: a[] = {2, 5, 3, 1, 7, 3, 4}
Output: 9
2->3->1->3

Minimum cost to reach the top of the floor by climbing stairs using Recursion:

This problem is an extension of problem: Climbing Stairs to reach at the top.

The thing is that we don’t need to go till the last element (since there are all positive numbers then we can avoid climbing to the last element so that we can reduce the cost).

Below is the implementation of the above idea:

C++

// C++ program to find the minimum
// cost required to reach the n-th floor
#include <bits/stdc++.h>
using namespace std;
 
// function to find the minimum cost
// to reach N-th floor
 
 
int minimumCost( int n , int cost[]){
    if(n == 0) return cost[0] ;
    if(n == 1) return cost[1] ;
   
     int top = min( minimumCost(n-1,cost) + cost[n] ,
                        minimumCost(n-2, cost)+ cost[n] );
   
      return top;
  
}
 
// Driver Code
int main()
{
    int a[] = { 16, 19, 10, 12, 18 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << minimumCost(n-2, a);
    return 0;
}

Java

/* Java code for finding minimum cost */
import java.io.*;
 
class GFG
{
 
  // function to find minimum cost
  public static int minimumCost(int n, int cost[]){
    if(n == 0) return cost[0] ;
    if(n == 1) return cost[1] ;
 
    int top = Math.min( minimumCost(n-1,cost) + cost[n] ,
                       minimumCost(n-2, cost)+ cost[n] );
    return top;
  }
  public static void main (String[] args) {
    int a[] = { 16, 19, 10, 12, 18 };
    int n = a.length;
    System.out.println(minimumCost(n-2,a));
  }
}
 
//  This code is contributed by rchandra

Python3

# Python3 program to find
# the minimum cost required
# to reach the n-th floor
 
# function to find the minimum
# cost to reach N-th floor
 
 
def minimumCost(cost, n):
    if n == 0:
        return cost[0]
    if n == 1:
        return cost[1]
    return min(minimumCost(cost, n-1) + cost[n], minimumCost(cost, n-2) + cost[n])
 
 
# Driver Code
if __name__ == "__main__":
    a = [16, 19, 10, 12, 18]
    n = len(a)
    print(minimumCost(a, n-2))

C#

/* C# code for finding minimum cost */
using System;
 
class GFG
{
 
  // function to find minimum cost
  public static int minimumCost(int n, int[] cost){
    if(n == 0) return cost[0] ;
    if(n == 1) return cost[1] ;
 
    int top = Math.Min( minimumCost(n-1,cost) + cost[n] ,
                       minimumCost(n-2, cost)+ cost[n] );
    return top;
  }
   
  static public void Main () {
    int[] a = { 16, 19, 10, 12, 18 };
    int n = a.Length;
    Console.Write(minimumCost(n-2,a));
  }
}
 
// This code is contributed by Pushpesh Raj.

Javascript

// Javascript program to find the minimum
// cost required to reach the n-th floor
 
// function to find the minimum cost
// to reach N-th floor
function minimumCost(n , cost){
    if(n == 0) return cost[0] ;
    if(n == 1) return cost[1] ;
 
    let top = Math.min( minimumCost(n-1,cost) + cost[n] ,
                        minimumCost(n-2, cost)+ cost[n] );
 
    return top;
 
}
 
// Driver Code
 
    let a = [ 16, 19, 10, 12, 18 ];
    let n = a.length;
    console.log(minimumCost(n-2, a));
     
// This code is contributed by Aman Kumar

Output

Time Complexity: O(N)
Auxiliary Space: O(1)

Minimum cost to reach the top of the floor by climbing stairs Dynamic Programming (Memoization):

We can store the result of calculated subproblems. So, that we don’t need to recalculate it again.

Below is the implementation of the above approach:

C++

// C++ program to find the minimum
// cost required to reach the n-th floor
#include <bits/stdc++.h>
using namespace std;
 
// function to find the minimum cost
// to reach N-th floor
int minimumCostMemoized(int n, vector<int>& cost,
                        vector<int>& dp)
{
    if (n == 0)
        return cost[0];
    if (n == 1)
        return cost[1];
    if (dp[n] != -1)
        return dp[n];
    dp[n] = min(
        minimumCostMemoized(n - 1, cost, dp) + cost[n],
        minimumCostMemoized(n - 2, cost, dp) + cost[n]);
    return dp[n];
}
 
int minCostClimbingStairs(vector<int>& cost)
{
    int n = cost.size();
    vector<int> dp(n + 1, -1);
    int ans = min(minimumCostMemoized(n - 2, cost, dp),
                  minimumCostMemoized(n - 1, cost, dp));
    return ans;
}
 
// Driver Code
int main()
{
    vector<int> a{ 10, 15, 20 };
    cout << minCostClimbingStairs(a);
    return 0;
}

Java

/*package whatever //do not write package name here */
 
import java.io.*;
import java.util.Arrays;
 
class GFG {
    // function to find the minimum cost
    // to reach N-th floor
    static int minimumCostMemoized(int n, int cost[],
                                   int dp[])
    {
        // base case
        if (n == 0)
            return dp[n] = cost[0];
        if (n == 1)
            return dp[n] = cost[1];
 
        if (dp[n] != -1)
            return dp[n];
 
        return dp[n]
            = Math.min(minimumCostMemoized(n - 1, cost, dp)
                           + cost[n],
                       minimumCostMemoized(n - 2, cost, dp)
                           + cost[n]);
    }
 
    static int minimumCost(int cost[], int n)
    {
        int dp[] = new int[n];
        Arrays.fill(dp, -1);
        int top = Math.min(
            minimumCostMemoized(n - 1, cost, dp),
            minimumCostMemoized(n - 2, cost, dp));
        return top;
    }
    public static void main(String[] args)
    {
        int a[] = { 10, 15, 20 };
        int n = a.length;
        System.out.println(minimumCost(a, n));
    }
}
 
// This code is contributed by rchandra.

Python3

# Python3 program to find
# the minimum cost required
# to reach the n-th floor
 
# function to find the minimum
# cost to reach N-th floor
 
 
def minimumCostMemoized(n, cost, dp):
    if n == 0:
        return cost[0]
    if n == 1:
        return cost[1]
    if dp[n] != -1:
        return dp[n]
    dp[n] = min(minimumCostMemoized(n - 1, cost, dp) + cost[n],
                minimumCostMemoized(n - 2, cost, dp) + cost[n])
    return dp[n]
 
 
def minCostClimbingStairs(cost):
    n = len(a)
    dp = [-1]*n
    ans = min(minimumCostMemoized(n - 2, cost, dp),
              minimumCostMemoized(n - 1, cost, dp))
    return ans
 
 
# Driver Code
if __name__ == "__main__":
    a = [16, 19, 10, 12, 18]
    print(minCostClimbingStairs(a))

C#

using System;
 
class GFG {
 
    // function to find the minimum cost
    // to reach N-th floor
    static int minimumCostMemoized(int n, int[] cost,
                                   int[] dp)
    {
 
        // base case
        if (n == 0)
            return cost[0];
        if (n == 1)
            return cost[1];
 
        if (dp[n] != -1)
            return dp[n];
 
        return dp[n]
            = Math.Min(minimumCostMemoized(n - 1, cost, dp)
                           + cost[n],
                       minimumCostMemoized(n - 2, cost, dp)
                           + cost[n]);
    }
 
    static int minimumCost(int[] cost, int n)
    {
        int[] dp = new int[n];
        for (int i = 0; i < n; i++)
            dp[i] = -1;
        int top = minimumCostMemoized(n - 2, cost, dp);
        return dp[n - 2];
    }
    public static void Main()
    {
        int[] a = { 16, 19, 10, 12, 18 };
        int n = a.Length;
        Console.WriteLine(minimumCost(a, n));
    }
}
 
// This code is contributed by Utkarsh

Javascript

// JS program to find the minimum
// cost required to reach the n-th floor
function minimumCostMemoized(n, cost, dp)
{
 
    // base case for reaching the 0th floor
    if (n === 0) {
        return cost[0];
    }
     
    // base case for reaching the 1st floor
    if (n === 1) {
        return cost[1];
    }
     
    // check if the subproblem is already solved
    if (dp[n] !== -1) {
        return dp[n];
    }
     
    // finding the minimum cost to reach the nth floor
    dp[n] = Math.min(minimumCostMemoized(n - 1, cost, dp) + cost[n], minimumCostMemoized(n - 2, cost, dp) + cost[n]);
    return dp[n];
}
 
function minCostClimbingStairs(cost) {
    var n = cost.length;
    var dp = new Array(n).fill(-1);
     
    // finding the minimum cost of reaching n-1 or n-2 floor
    var ans = Math.min(minimumCostMemoized(n - 2, cost, dp), minimumCostMemoized(n - 1, cost, dp));
    return ans;
}
 
// Driver Code
 
    var a = [16, 19, 10, 12, 18];
    console.log(minCostClimbingStairs(a));
 
// This code is contributed by lokeshpotta20.

Output

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

Minimum cost to reach the top of the floor by climbing stairs using Dynamic Programming (Tabulation):

Let dp[i] be the cost to climb the i-th staircase to from 0-th or 1-th step. Hence dp[i] = cost[i] + min(dp[i-1], dp[i-2]). Since dp[i-1] and dp[i-2] are needed to compute the cost of travelling from i-th step, a bottom-up approach can be used to solve the problem. The answer will be the minimum cost of reaching n-1^th stair and n-2^th stair. Compute the dp[] array in a bottom-up manner.

Below is the implementation of the above approach.

C++

// C++ program to find the minimum
// cost required to reach the n-th floor
#include <bits/stdc++.h>
using namespace std;
 
// function to find the minimum cost
// to reach N-th floor
int minimumCost(int cost[], int n)
{
    // declare an array
    int dp[n];
 
    // base case
    if (n == 1)
        return cost[0];
 
    // initially to climb till 0-th
    // or 1th stair
    dp[0] = cost[0];
    dp[1] = cost[1];
 
    // iterate for finding the cost
    for (int i = 2; i < n; i++) {
        dp[i] = min(dp[i - 1], dp[i - 2]) + cost[i];
    }
 
    // return the minimum
    return min(dp[n - 2],dp[n-1]);
}
 
// Driver Code
int main()
{
    int a[] = { 16, 19, 10, 12, 18 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << minimumCost(a, n);
    return 0;
}

Java

// Java program to find the 
// minimum cost required to
// reach the n-th floor
import java.io.*;
import java.util.*;
 
class GFG
{
// function to find 
// the minimum cost
// to reach N-th floor
static int minimumCost(int cost[], 
                       int n)
{
    // declare an array
    int dp[] = new int[n];
 
    // base case
    if (n == 1)
        return cost[0];
 
    // initially to 
    // climb till 0-th
    // or 1th stair
    dp[0] = cost[0];
    dp[1] = cost[1];
 
    // iterate for finding the cost
    for (int i = 2; i < n; i++)
    {
        dp[i] = Math.min(dp[i - 1], 
                         dp[i - 2]) + cost[i];
    }
 
    // return the minimum
    return Math.min(dp[n - 2], 
                    dp[n - 1]);
}
 
// Driver Code
public static void main(String args[])
{
    int a[] = { 16, 19, 10, 12, 18 };
    int n = a.length;
    System.out.print(minimumCost(a, n));
}
}

Python3

# Python3 program to find 
# the minimum cost required 
# to reach the n-th floor
 
# function to find the minimum 
# cost to reach N-th floor
def minimumCost(cost, n):
 
    # declare an array
    dp = [None]*n
 
    # base case
    if n == 1:
        return cost[0]
 
    # initially to climb 
    # till 0-th or 1th stair
    dp[0] = cost[0]
    dp[1] = cost[1]
 
    # iterate for finding the cost
    for i in range(2, n): 
        dp[i] = min(dp[i - 1], 
                    dp[i - 2]) + cost[i]
 
    # return the minimum
    return min(dp[n - 2], dp[n - 1])
 
# Driver Code
if __name__ == "__main__":
    a = [16, 19, 10, 12, 18 ]
    n = len(a)
    print(minimumCost(a, n))
 
# This code is contributed 
# by ChitraNayal

C#

// C# program to find the 
// minimum cost required to
// reach the n-th floor
using System;
 
class GFG
{
// function to find 
// the minimum cost
// to reach N-th floor
static int minimumCost(int[] cost, 
                       int n)
{
    // declare an array
    int []dp = new int[n];
 
    // base case
    if (n == 1)
        return cost[0];
 
    // initially to 
    // climb till 0-th
    // or 1th stair
    dp[0] = cost[0];
    dp[1] = cost[1];
 
    // iterate for finding the cost
    for (int i = 2; i < n; i++)
    {
        dp[i] = Math.Min(dp[i - 1], 
                         dp[i - 2]) + cost[i];
    }
 
    // return the minimum
    return Math.Min(dp[n - 2], 
                    dp[n - 1]);
}
 
// Driver Code
public static void Main()
{
    int []a = { 16, 19, 10, 12, 18 };
    int n = a.Length;
    Console.WriteLine(minimumCost(a, n));
}
}
 
// This code is contributed
// by Subhadeep

Javascript

<script>
 
// Javascript program to find the
// minimum cost required to
// reach the n-th floor
 
     
    // function to find
// the minimum cost
// to reach N-th floor    
    function minimumCost(cost,n)
    {
        // declare an array
    let dp = new Array(n);
  
    // base case
    if (n == 1)
        return cost[0];
  
    // initially to
    // climb till 0-th
    // or 1th stair
    dp[0] = cost[0];
    dp[1] = cost[1];
  
    // iterate for finding the cost
    for (let i = 2; i < n; i++)
    {
        dp[i] = Math.min(dp[i - 1],
                         dp[i - 2]) + cost[i];
    }
  
    // return the minimum
    return Math.min(dp[n - 2],
                    dp[n - 1]);
    }
     
    // Driver Code
    let a=[16, 19, 10, 12, 18 ];
    let n = a.length;
    document.write(minimumCost(a, n));
     
 
// This code is contributed by rag2127
 
</script>

PHP

<?php
// PHP program to find the 
// minimum cost required 
// to reach the n-th floor
 
// function to find the minimum 
// cost to reach N-th floor
function minimumCost(&$cost, $n)
{
    // declare an array
 
    // base case
    if ($n == 1)
        return $cost[0];
 
    // initially to climb 
    // till 0-th or 1th stair
    $dp[0] = $cost[0];
    $dp[1] = $cost[1];
 
    // iterate for finding 
    // the cost
    for ($i = 2; $i < $n; $i++) 
    {
        $dp[$i] = min($dp[$i - 1], 
                      $dp[$i - 2]) + 
                      $cost[$i];
    }
 
    // return the minimum
    return min($dp[$n - 2], 
               $dp[$n - 1]);
}
 
// Driver Code
$a = array(16, 19, 10, 12, 18);
$n = sizeof($a);
echo(minimumCost($a, $n));
     
// This code is contributed
// by Shivi_Aggarwal
?>

Output

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

Minimum cost to reach the top of the floor by climbing stairs using Dynamic Programming (Space Optimized):

Instead of using dp[] array for memoizing the cost, use two-variable dp1 and dp2. Since the cost of reaching the last two stairs is required only, use two variables and update them by swapping when one stair is climbed.

Below is the implementation of the above approach:

C++

// C++ program to find the minimum
// cost required to reach the n-th floor
// space-optimized solution
#include <bits/stdc++.h>
using namespace std;
 
// function to find the minimum cost
// to reach N-th floor
int minimumCost(int cost[], int n)
{
    int dp1 = 0, dp2 = 0;
 
    // traverse till N-th stair
    for (int i = 0; i < n; i++) {
        int dp0 = cost[i] + min(dp1, dp2);
 
        // update the last two stairs value
        dp2 = dp1;
        dp1 = dp0;
    }
    // dp2 gives the cost if started climbing from index 1
    // and dp1 from index 0
    return min(dp2, dp1);
}
// Driver Code
int main()
{
    int a[] = { 2, 5, 3, 1, 7, 3, 4 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << minimumCost(a, n);
    return 0;
}

Java

// Java program to find the 
// minimum cost required to 
// reach the n-th floor 
// space-optimized solution
import java.io.*;
import java.util.*;
 
class GFG
{
// function to find 
// the minimum cost
// to reach N-th floor
static int minimumCost(int cost[], int n)
{
    int dp1 = 0, dp2 = 0;
 
    // traverse till N-th stair
    for (int i = 0; i < n; i++) 
    {
        int dp0 = cost[i] + 
                  Math.min(dp1, dp2);
 
        // update the last 
        // two stairs value
        dp2 = dp1;
        dp1 = dp0;
    }
    return Math.min(dp1, dp2);
}
 
// Driver Code
public static void main(String args[])
{
    int a[] = { 2, 5, 3, 1, 7, 3, 4 };
    int n = a.length;
    System.out.print(minimumCost(a, n));
}
}

Python3

# Python3 program to find 
# the minimum cost required 
# to reach the n-th floor
# space-optimized solution
 
# function to find the minimum 
# cost to reach N-th floor
def minimumCost(cost, n):
 
    dp1 = 0
    dp2 = 0
 
    # traverse till N-th stair
    for i in range(n):
        dp0 = cost[i] + min(dp1, dp2)
 
        # update the last
        # two stairs value
        dp2 = dp1
        dp1 = dp0
    return min(dp1, dp2)
 
# Driver Code
if __name__ == "__main__":
    a = [ 2, 5, 3, 1, 7, 3, 4 ]
    n = len(a)
    print(minimumCost(a, n))
     
# This code is contributed
# by ChitraNayal

C#

// C# program to find the 
// minimum cost required to 
// reach the n-th floor 
// space-optimized solution
using System;
 
class GFG
{
// function to find 
// the minimum cost
// to reach N-th floor
static int minimumCost(int[] cost,
                       int n)
{
    int dp1 = 0, dp2 = 0;
 
    // traverse till N-th stair
    for (int i = 0; i < n; i++) 
    {
        int dp0 = cost[i] + 
                  Math.Min(dp1, dp2);
 
        // update the last 
        // two stairs value
        dp2 = dp1;
        dp1 = dp0;
    }
    return Math.Min(dp1, dp2);
}
 
// Driver Code
public static void Main()
{
    int[] a = { 2, 5, 3, 1, 7, 3, 4 };
    int n = a.Length;
    Console.Write(minimumCost(a, n));
}
}
 
// This code is contributed
// by ChitraNayal

Javascript

<script>
// Javascript program to find the
// minimum cost required to
// reach the n-th floor
// space-optimized solution
     
    // function to find
// the minimum cost
// to reach N-th floor
    function minimumCost(cost,n)
    {
        let dp1 = 0, dp2 = 0;
  
    // traverse till N-th stair
    for (let i = 0; i < n; i++)
    {
        let dp0 = cost[i] +
                  Math.min(dp1, dp2);
  
        // update the last
        // two stairs value
        dp2 = dp1;
        dp1 = dp0;
    }
    return Math.min(dp1, dp2);
    }
     
    // Driver Code
    let a=[2, 5, 3, 1, 7, 3, 4 ];
    let n = a.length;
    document.write(minimumCost(a, n));
     
 
// This code is contributed by avanitrachhadiya2155
</script>

PHP

<?php
// PHP program to find the 
// minimum cost required to 
// reach the n-th floor 
// space-optimized solution
 
// function to find the minimum 
// cost to reach N-th floor
function minimumCost(&$cost, $n)
{
    $dp1 = 0;
    $dp2 = 0;
 
    // traverse till N-th stair
    for ($i = 0; $i < $n; $i++) 
    {
        $dp0 = $cost[$i] +
               min($dp1, $dp2);
 
        // update the last 
        // two stairs value
        $dp2 = $dp1;
        $dp1 = $dp0;
    }
    return min($dp1, $dp2);
}
// Driver Code
$a = array(2, 5, 3, 1, 7, 3, 4);
$n = sizeof($a);
echo (minimumCost($a, $n));
 
// This code is contributed
// by Shivi_Aggarwal
?>

Output

Time Complexity: O(N)
Auxiliary Space: O(1)

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Minimum cost to reach the top of the floor by climbing stairs

Minimum cost to reach the top of the floor by climbing stairs using Recursion:

C++

Java

Python3

C#

Javascript

Minimum cost to reach the top of the floor by climbing stairs Dynamic Programming (Memoization):

C++

Java

Python3

C#

Javascript

Minimum cost to reach the top of the floor by climbing stairs using Dynamic Programming (Tabulation):

C++

Java

Python3

C#

Javascript

PHP

Minimum cost to reach the top of the floor by climbing stairs using Dynamic Programming (Space Optimized):

C++

Java

Python3

C#

Javascript

PHP

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