April day 12 done

problems/1046/main.rs

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+pub fn last_stone_weight(mut stones: Vec<i32>) -> i32 {
+    if stones.len() == 1 {
+        return *stones.get(0).unwrap();
+    } else if stones.len() == 0 {
+        return 0;
+    }
+    // sort
+    stones.sort_by(|a, b| b.cmp(a));
+    // remove the max twice
+    let m1 = stones.remove(0);
+    let m2 = stones.remove(0);
+    // add result, if any
+    if m1 != m2 {
+        stones.push(m1 - m2);
+    }
+    // go again
+    return last_stone_weight(stones);
+}
+
+pub fn main() {
+    println!("Expected: 1");
+    println!("Got: {}", last_stone_weight(vec![2, 7, 4, 1, 8, 1]));
+}

problems/1046/problem.txt

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+We have a collection of stones, each stone has a positive integer weight.
+
+Each turn, we choose the two heaviest stones and smash them together.  Suppose the stones have weights x and y with x <= y.  The result of this smash is:
+
+    If x == y, both stones are totally destroyed;
+    If x != y, the stone of weight x is totally destroyed, and the stone of weight y has new weight y-x.
+
+At the end, there is at most 1 stone left.  Return the weight of this stone (or 0 if there are no stones left.)
+
+ 
+
+Example 1:
+
+Input: [2,7,4,1,8,1]
+Output: 1
+Explanation: 
+We combine 7 and 8 to get 1 so the array converts to [2,4,1,1,1] then,
+we combine 2 and 4 to get 2 so the array converts to [2,1,1,1] then,
+we combine 2 and 1 to get 1 so the array converts to [1,1,1] then,
+we combine 1 and 1 to get 0 so the array converts to [1] then that's the value of last stone.
+
+ 
+
+Note:
+
+    1 <= stones.length <= 30
+    1 <= stones[i] <= 1000
+

problems/1046/run.sh

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+#!/bin/bash
+
+rustc main.rs
+./main
+rm main