add no.3761:

use hash map

Author: clingoram

javascript/LeetCode/Array/3761.js

@@ -0,0 +1,44 @@
+/**
+ * 3761. Minimum Absolute Distance Between Mirror Pairs
+ * 
+ * mirror pair = indices(i,j)
+ * reverse(nums[i] === nums[j]) 若數字前面為0，則省略0
+ * 回傳最小mirror pair絕對距離 abs(i - j)，若無，回傳-1
+ * 
+ * @param {number[]} nums
+ * @return {number}
+ */
+var minMirrorPairDistance = function(nums) {
+  /**
+   * 陣列元素兩個為一組(i,j)，每個元素反轉後跟下一個元素比較是否一致。若一致 abs(index i - index j)，取最小結果
+   */
+  // 反轉數字
+  function reverseNum(x){
+    let y = 0;
+    while(x > 0){
+      y = y * 10 + (x % 10);
+      x = Math.floor(x / 10);
+    }
+    return y;
+  }
+
+  let map = new Map();
+  let ans = nums.length + 1;
+  for(let i = 0;i < nums.length;i++){
+    if(map.has(nums[i])){
+      ans = Math.min(ans,i - map.get(nums[i]));
+    }
+    map.set(reverseNum(nums[i]),i);
+  }
+  return ans === nums.length + 1 ? -1 : ans;
+};
+let nums = [12,21,45,33,54]
+/*
+Output: 1
+Explanation:
+The mirror pairs are:
+(0, 1) since reverse(nums[0]) = reverse(12) = 21 = nums[1], giving an absolute distance abs(0 - 1) = 1.
+(2, 4) since reverse(nums[2]) = reverse(45) = 54 = nums[4], giving an absolute distance abs(2 - 4) = 2.
+The minimum absolute distance among all pairs is 1.
+*/
+console.log(minMirrorPairDistance(nums));

javascript/index.js

@@ -1408,6 +1408,7 @@ var getLeastFrequentDigit = function(n) {
      * solution 1. Hash table
      * solution 2. Array
      */
+
     // solution 1.
     // Hash table
     // let nSplitToStr = n.toString().split("");
@@ -1416,15 +1417,18 @@ var getLeastFrequentDigit = function(n) {
     // for(let i = 0;i < nSplitToStr.length;++i) {
     //   map.has(nSplitToStr[i]) ? map.set(nSplitToStr[i],map.get(nSplitToStr[i]) + 1) : map.set(nSplitToStr[i],1);
     // }
+    // 不斷比較minFreq和value哪個最小，因此minFreq值會一直更新
     // for(const [key,value] of map){
-    //   minFreq = Math.min(minFreq, value);
+    //     minFreq = Math.min(minFreq,value);
     // }
-    // for(const [key,value] of map) {
-    //   if(value === minFreq){
-    //     result = Math.min(result,key);
-    //   }
+    // for(const [key,value] of map){
+    //     // 最小的value = minFreq
+    //     if(value === minFreq){
+    //         // 比較result和key哪個最小，key = 元素
+    //         result = Math.min(result,key);
+    //     }
     // }
-    // return result;
+    // return reuslt;
 
     // solution 2.
     // Array.
@@ -1439,7 +1443,7 @@ Output: 2
 Explanation:
 The least frequent digits in n are 7, 2, and 5; each appears only once.
 */
-console.log(getLeastFrequentDigit(n));
+// console.log(getLeastFrequentDigit(n));
 
 /**
  * 3488. Closest Equal Element Queries
@@ -1457,11 +1461,13 @@ console.log(getLeastFrequentDigit(n));
 var solveQueries = function(nums, queries) {
 
 };
-let nums = [1,3,1,4,1,3,2], queries = [0,3,5];
+// let nums = [1,3,1,4,1,3,2], queries = [0,3,5];
 /*
 Output: [2,-1,3]
 Explanation:
 Query 0: The element at queries[0] = 0 is nums[0] = 1. The nearest index with the same value is 2, and the distance between them is 2.
 Query 1: The element at queries[1] = 3 is nums[3] = 4. No other index contains 4, so the result is -1.
 Query 2: The element at queries[2] = 5 is nums[5] = 3. The nearest index with the same value is 1, and the distance between them is 3 (following the circular path: 5 -> 6 -> 0 -> 1).
-*/
+*/
+
+