Maximum Gap - Radix Sort [JS]
Description Solution: Radix Sort Get the max number from nums and count the number of digits it has. Loop for maxDigits number of times: arrange each nums[i] into its bucket based on the current digit prefix sum/accumulate indices in count place each nums[i] to next based on the accumulated index in count reassign nums to next d = digits <= 10, n = length of nums, b = base = 10 Time Complexity: O(d(n + b)) = O(n) 310ms Space Complexity: O(n) 67.2MB var maximumGap = function ( nums ) { let n = nums.length, max = Math .max(...nums), maxDigits = getDigits(max); let power = 1 ; for ( let k = 0 ; k < maxDigits; k++) { let next = Array (n), count = Array ( 10 ).fill( 0 ); for ( let i = 0 ; i < n; i++) { let digit = Math .floor(nums[i] / power) % 10 ; count[digit]++; // counting sort based on current digit } for ( let i = 1 ; i < 10 ; i++) count[i] += count[i - 1 ]; // prefix sum for ( let i = n - 1 ; i >= 0 ; i--) {...