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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0001_0100.s0070_climbing_stairs;

// #Easy #Top_100_Liked_Questions #Top_Interview_Questions #Dynamic_Programming #Math #Memoization
// #Algorithm_I_Day_12_Dynamic_Programming #Dynamic_Programming_I_Day_2
// #Level_1_Day_10_Dynamic_Programming #Udemy_Dynamic_Programming #Big_O_Time_O(n)_Space_O(n)
// #2023_08_11_Time_0_ms_(100.00%)_Space_39.2_MB_(71.51%)

/**
 * 70 - Climbing Stairs\.
 *
 * Easy
 *
 * You are climbing a staircase. It takes `n` steps to reach the top.
 *
 * Each time you can either climb `1` or `2` steps. In how many distinct ways can you climb to the top?
 *
 * **Example 1:**
 *
 * **Input:** n = 2
 *
 * **Output:** 2
 *
 * **Explanation:** There are two ways to climb to the top. 1. 1 step + 1 step 2. 2 steps 
 *
 * **Example 2:**
 *
 * **Input:** n = 3
 *
 * **Output:** 3
 *
 * **Explanation:** There are three ways to climb to the top. 1. 1 step + 1 step + 1 step 2. 1 step + 2 steps 3. 2 steps + 1 step 
 *
 * **Constraints:**
 *
 * *   `1 <= n <= 45`
**/
public class Solution {
    public int climbStairs(int n) {
        if (n < 2) {
            return n;
        }
        int[] cache = new int[n];
        // creating a cache or DP to store the result
        // so that we dont have to iterate multiple times
        // for the same values;

        // for 0 and 1 the result array i.e cache values would be 1 and 2
        // in loop we are just getting ith values i.e 5th step values from
        // i-1 and i-2 which are 4th step and 3rd step values.
        cache[0] = 1;
        cache[1] = 2;
        for (int i = 2; i < n; i++) {
            cache[i] = cache[i - 1] + cache[i - 2];
        }
        return cache[n - 1];
    }
}