'Triangle' solns

Author: joewnga

leetcode/triangle.rs

@@ -0,0 +1,80 @@
+/// 
+/// Problem: Triangle
+/// 
+/// Given a triangle array, return the minimum path sum from top to bottom.
+/// 
+/// For each step, you may move to an adjacent number of the row below. More formally, 
+/// if you are on index i on the current row, you may move to either index i or index i + 1 
+/// on the next row.
+/// 
+/// Example 1:
+/// Input: triangle = [[2],[3,4],[6,5,7],[4,1,8,3]]
+/// Output: 11
+/// Explanation: The triangle looks like:
+///    2
+///   3 4
+///  6 5 7
+/// 4 1 8 3
+/// The minimum path sum from top to bottom is 2 + 3 + 5 + 1 = 11.
+/// 
+/// Example 2:
+/// Input: triangle = [[-10]]
+/// Output: -10
+/// 
+/// Constraints:
+/// 1 <= triangle.length <= 200
+/// triangle[0].length == 1
+/// triangle[i].length == triangle[i - 1].length + 1
+/// -10^4 <= triangle[i][j] <= 10^4
+/// 
+/// Follow up: Could you do this using only O(n) extra space, where n is the total number
+/// of rows in the triangle?
+/// 
+
+// # Solution 1
+// Time complexity: O(n²) 
+// Space complexity: O(n²) 
+
+
+impl Solution {
+    pub fn minimum_total(triangle: Vec<Vec<i32>>) -> i32 {
+        let n = triangle.len();
+       
+       let mut dp = vec![vec![0; n]; n];
+       
+       for j in 0..n {
+           dp[n - 1][j] = triangle[n - 1][j];
+       }
+       
+       for i in (0..n-1).rev() {
+           for j in 0..=i {
+            
+               dp[i][j] = triangle[i][j] + std::cmp::min(dp[i + 1][j], dp[i + 1][j + 1]);
+           }
+       }
+       
+       dp[0][0]
+    }
+}
+
+
+// # Solution 2:
+// Time complexity: O(n²) 
+// Space complexity: O(n)
+
+impl Solution {
+    pub fn minimum_total(triangle: Vec<Vec<i32>>) -> i32 {
+       let n = triangle.len();
+       
+       let mut dp = triangle[n - 1].clone();
+       
+       for i in (0..n-1).rev() {
+           for j in 0..=i {
+               
+               dp[j] = triangle[i][j] + std::cmp::min(dp[j], dp[j + 1]);
+           }
+       }
+       
+       dp[0]
+    }
+}