leetcode-weekly-237

Author: OneCodeMonkey

Diff for: contest/leetcode_week_237/leetcode_1832_Check_if_the_Sentence_Is_Pangram.java

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+// AC
+// Runtime: 5 ms, faster than 25.00% of Java online submissions for Check if the Sentence Is Pangram.
+// Memory Usage: 39 MB, less than 25.00% of Java online submissions for Check if the Sentence Is Pangram.
+class Solution {
+    public boolean checkIfPangram(String sentence) {
+        int size = sentence.length();
+        Map<String, Integer> record = new HashMap<>();
+        for(int i = 0; i < size; i++) {
+            String temp = String.valueOf(sentence.charAt(i));
+            record.merge(temp, 1, Integer::sum);
+        }
+        return record.size() == 26;
+    }
+}

Diff for: contest/leetcode_week_237/leetcode_1833_Maximum_Ice_Cream_Bars.java

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+// AC:
+// Runtime: 28 ms, faster than 100.00% of Java online submissions for Maximum Ice Cream Bars.
+// Memory Usage: 55.9 MB, less than 25.00% of Java online submissions for Maximum Ice Cream Bars.
+class Solution {
+    public int maxIceCream(int[] costs, int coins) {
+        Arrays.sort(costs);
+        int sum = 0;
+        for(int i = 0; i < costs.length; i++) {
+            sum += costs[i];
+            if (sum == coins) {
+                return i + 1;
+            }
+            if (sum > coins) {
+                if (sum == 0) {
+                    return 0;
+                } else {
+                    return i;
+                }
+            }
+        }
+        return costs.length;
+    }
+}

Diff for: contest/leetcode_week_237/leetcode_1834_Single-Threaded_CPU.java

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+// AC:
+// Runtime: 103 ms, faster than 57.14% of Java online submissions for Single-Threaded CPU.
+// Memory Usage: 96.7 MB, less than 14.29% of Java online submissions for Single-Threaded CPU.
+// 思路：用优先级队列，模拟当前 CPU 可选的任务池，
+// 
+class Solution {
+    public int[] getOrder(int[][] tasks) {
+        int taskCount = tasks.length;
+        int[] ret = new int[taskCount];
+        int[][] taskRecord = new int[taskCount][3];
+        for(int i = 0; i < taskCount; i++) {
+            taskRecord[i][0] = i;
+            taskRecord[i][1] = tasks[i][0];
+            taskRecord[i][2] = tasks[i][1];
+        }
+        Arrays.sort(taskRecord, (a, b) -> (a[1] - b[1]));
+        // 按执行时间越短排越前
+        PriorityQueue<int[]> queue = new PriorityQueue<int[]>((a, b) -> a[2] == b[2] ? a[0] - b[0] : a[2] - b[2]);
+        int time = 0, step = 0, timeStep = 0;
+        while (step < taskCount) {
+            while (timeStep < taskCount && taskRecord[timeStep][1] <= time) {
+                queue.offer(taskRecord[timeStep++]);
+            }
+            if (queue.isEmpty()) {
+                time = taskRecord[timeStep][1];
+                continue;
+            }
+            int[] curTask = queue.poll();
+            ret[step++] = curTask[0];
+            time += curTask[2];
+        }
+
+        return ret;
+    }
+}

Diff for: contest/leetcode_week_237/leetcode_1835_Find_XOR_Sum_of_All_Pairs_Bitwise_AND.java

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+// AC:
+// Runtime: 1 ms, faster than 100.00% of Java online submissions for Find XOR Sum of All Pairs Bitwise AND.
+// Memory Usage: 52.3 MB, less than 25.00% of Java online submissions for Find XOR Sum of All Pairs Bitwise AND.
+// 
+// 思路：这个运算场景满足结合律： (a1 ^ a2) & (b1 ^ b2) = (a1 & b1) ^ (a1 & b2) ^ (a2 & b1) ^ (a2 & b2)
+// 复杂度：T：O(m + n), S:O(1)
+class Solution {
+    public int getXORSum(int[] arr1, int[] arr2) {
+        int ret1 = 0, ret2 = 0;
+        for(int i: arr1) {
+            ret1 ^= i;
+        }
+        for(int i: arr2) {
+            ret2 ^= i;
+        }
+        return ret1 & ret2;
+    }
+}