Created
February 6, 2026 20:20
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You are given an integer array nums and an integer k. An array is considered balanced if the value of its maximum element is at most k times the minimum element. You may remove any number of elements from nums without making it empty. Retur
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| /** | |
| * @param {number[]} nums | |
| * @param {number} k | |
| * @return {number} | |
| */ | |
| var minRemoval = function(nums, k) { | |
| // Step 1: Sort the array so window are contiguous and ordered | |
| nums.sort((a, b) => a - b); | |
| let n = nums.length; | |
| let i = 0; // left pointer (minimum in window) | |
| let maxWindow = 1; // at least one element is always balanced | |
| // Step 2: Slide the right pointer j across the array | |
| for (let j = 0; j < n; j++) { | |
| // Step 3: If window becomes invalid, move i forward | |
| // Condition for invalid window: | |
| // nums[j] > k * nums[i] | |
| while (nums[j] > k * nums[i]) { | |
| i++; // shrink from the left until valid | |
| } | |
| // Step 4: Update the largest valid window size | |
| maxWindow = Math.max(maxWindow, j - i + 1); | |
| } | |
| // Step 5: Minimum removals = total - largest balanced window | |
| return n - maxWindow; | |
| }; |
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