First hard problem done

problems/1402/main.rs

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+use std::cmp;
+
+pub fn max_satisfaction(mut satisfaction: Vec<i32>) -> i32 {
+    // First we want to sort the dishes, because the maximum like-time will come
+    // when the last dishes are highest in number - then we just need to work out which ones to cull
+    // there is one case that can probably be optimised: all negative (cook nothing)
+    // but it might be easier to just build that into the starting logic
+    satisfaction.sort_by(|a, b| b.cmp(a));
+    let mut max = 0;
+    let mut sum = 0; // sum will count how much will be added by multiplying the end of the arr
+    // Now loop backwards over the array seeing if it makes things better to add dishes
+    // Note: The array has been reversed, so it's actually going forwards, trippy right?
+    for dish in satisfaction {
+        let total = max + dish + sum; // This is the total if this dish was added
+        if total > max {
+            sum = sum + dish;
+            max = total;
+        } else {
+            return max; // we can't do better, stop here
+        }
+    }
+    return cmp::max(0, max);
+}
+
+pub fn main() {
+    println!("Expected: 20");
+    println!("Got: {}", max_satisfaction(vec![4, 3, 2]));
+}

problems/1402/problem.txt

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+A chef has collected data on the satisfaction level of his n dishes. Chef can cook any dish in 1 unit of time.
+
+Like-time coefficient of a dish is defined as the time taken to cook that dish including previous dishes multiplied by its satisfaction level  i.e.  time[i]*satisfaction[i]
+
+Return the maximum sum of Like-time coefficient that the chef can obtain after dishes preparation.
+
+Dishes can be prepared in any order and the chef can discard some dishes to get this maximum value.
+
+ 
+
+Example 1:
+
+Input: satisfaction = [-1,-8,0,5,-9]
+Output: 14
+Explanation: After Removing the second and last dish, the maximum total Like-time coefficient will be equal to (-1*1 + 0*2 + 5*3 = 14). Each dish is prepared in one unit of time.
+
+Example 2:
+
+Input: satisfaction = [4,3,2]
+Output: 20
+Explanation: Dishes can be prepared in any order, (2*1 + 3*2 + 4*3 = 20)
+
+Example 3:
+
+Input: satisfaction = [-1,-4,-5]
+Output: 0
+Explanation: People don't like the dishes. No dish is prepared.
+
+Example 4:
+
+Input: satisfaction = [-2,5,-1,0,3,-3]
+Output: 35
+
+
+Constraints:
+
+    n == satisfaction.length
+    1 <= n <= 500
+    -10^3 <= satisfaction[i] <= 10^3

problems/1402/run.sh

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