add 2654

Author: clingoram

javascript/LeetCode/math/2654.js

@@ -0,0 +1,59 @@
+/**
+ * 2654. Minimum Number of Operations to Make All Array Elements Equal to 1
+ * 
+ * GCD = 最大公因數」（Greatest Common Divisor）
+ * 陣列元素可操作數次，選任一index i (0 <= i < nums.length - 1)並取得nums[i] or nums[i + 1]的GCD
+ * 若可讓每個元素都=1的話，回傳1，否則-1
+ * 
+ * @param {number[]} nums
+ * @return {number}
+ */
+var minOperations = function(nums) {
+  let g = 0,nums1 = 0;
+
+  function mygcd(x,y){
+    // 求x & y 的最大公約數
+    // x & y 可能是很大的數
+    if (y === 0) {
+      return x;
+    }
+    return mygcd(y, x % y);
+  }
+
+  for(const a of nums) {
+    if(a === 1) {
+      nums1++;
+    }
+    g = mygcd(g,a);
+  }
+  if(nums1 > 0) {
+    return nums.length - nums1;
+  }
+  if(g > 1) {
+    return -1;
+  }
+
+  let minLen = nums.length;
+  for (let i = 0; i < nums.length; ++i) {
+    let currentGcd = 0;
+    for (let j = i; j < nums.length; ++j) {
+      currentGcd = mygcd(currentGcd, nums[j]);
+      if(currentGcd === 1) {
+        minLen = Math.min(minLen, j - i + 1);
+        break;
+      }
+    }
+  }
+  return minLen + nums.length - 2;
+
+};
+// let nums = [2,6,3,4];
+//  4
+// Explanation: We can do the following operations:
+// - Choose index i = 2 and replace nums[2] with gcd(3,4) = 1. Now we have nums = [2,6,1,4].
+// - Choose index i = 1 and replace nums[1] with gcd(6,1) = 1. Now we have nums = [2,1,1,4].
+// - Choose index i = 0 and replace nums[0] with gcd(2,1) = 1. Now we have nums = [1,1,1,4].
+// - Choose index i = 2 and replace nums[3] with gcd(1,4) = 1. Now we have nums = [1,1,1,1].
+let nums = [1,1];
+// 0
+console.log(minOperations(nums));

javascript/index.js

@@ -1079,9 +1079,10 @@ var findXSum = function(nums, k, x) {
   // }
 
 };
-let nums = [1,1,2,2,3,4,2,3], k = 6, x = 2;
+// let nums = [1,1,2,2,3,4,2,3], k = 6, x = 2;
 // [6,10,12]
 // For subarray [1, 1, 2, 2, 3, 4], only elements 1 and 2 will be kept in the resulting array. Hence, answer[0] = 1 + 1 + 2 + 2.
 // For subarray [1, 2, 2, 3, 4, 2], only elements 2 and 4 will be kept in the resulting array. Hence, answer[1] = 2 + 2 + 2 + 4. Note that 4 is kept in the array since it is bigger than 3 and 1 which occur the same number of times.
 // For subarray [2, 2, 3, 4, 2, 3], only elements 2 and 3 are kept in the resulting array. Hence, answer[2] = 2 + 2 + 2 + 3 + 3.
-console.log(findXSum(nums,k,x));
+// console.log(findXSum(nums,k,x));
+