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| 1 | +/** |
| 2 | + * https://leetcode.com/problems/maximum-number-of-removable-characters/ |
| 3 | + * |
| 4 | + * Brute force |
| 5 | + * Time O(removable.length * s.length) | Space O(n) |
| 6 | + * @param {string} s |
| 7 | + * @param {string} p |
| 8 | + * @param {number[]} removable |
| 9 | + * @return {number} |
| 10 | + */ |
| 11 | +var maximumRemovals1 = function(s, p, removable) { |
| 12 | + let k = 0; |
| 13 | + // removable.reverse(); |
| 14 | + s = s.split(''); |
| 15 | + p = p.split(''); |
| 16 | + |
| 17 | + for (let i = 0; i < removable.length; i++) { |
| 18 | + s[removable[i]] = -1; |
| 19 | + if (isSubSet1(s, p)) { |
| 20 | + k++; |
| 21 | + continue; |
| 22 | + } |
| 23 | + return k; |
| 24 | + } |
| 25 | + |
| 26 | + return k; |
| 27 | +}; |
| 28 | + |
| 29 | +// helper function. |
| 30 | +function isSubSet1(s, p) { |
| 31 | + let i = 0; |
| 32 | + let j = 0; |
| 33 | + |
| 34 | + while (i < s.length && j < p.length) { |
| 35 | + if (s[i] === p[j]) { |
| 36 | + i++; |
| 37 | + j++; |
| 38 | + } else { |
| 39 | + i++; |
| 40 | + } |
| 41 | + } |
| 42 | + |
| 43 | + return j === p.length; |
| 44 | +} |
| 45 | + |
| 46 | + |
| 47 | +/** |
| 48 | + * |
| 49 | + * Binary Search |
| 50 | + * n = length of string, k = length of removable |
| 51 | + * Time O(log(k)*n) | Space O(1) |
| 52 | + * @param {string} s |
| 53 | + * @param {string} p |
| 54 | + * @param {number[]} removable |
| 55 | + * @return {number} |
| 56 | + */ |
| 57 | +var maximumRemovals = function(s, p, removable) { |
| 58 | + |
| 59 | + let left = 0; |
| 60 | + let right = removable.length - 1; |
| 61 | + let k = 0; |
| 62 | + |
| 63 | + while (left <= right) { |
| 64 | + const mid = (left + right) >> 1; |
| 65 | + const hash = new Set(removable.slice(0, mid + 1)); |
| 66 | + |
| 67 | + if (isSubSet(hash, s, p)) { |
| 68 | + k = Math.max(k, mid + 1); |
| 69 | + left = mid + 1; |
| 70 | + continue; |
| 71 | + } |
| 72 | + |
| 73 | + right = mid - 1; |
| 74 | + } |
| 75 | + |
| 76 | + return k; |
| 77 | +}; |
| 78 | + |
| 79 | +// helper function. |
| 80 | +function isSubSet(hash, s, p) { |
| 81 | + let i = 0; |
| 82 | + let j = 0; |
| 83 | + |
| 84 | + while (i < s.length && j < p.length) { |
| 85 | + if (s[i] === p[j] && !hash.has(i)) { |
| 86 | + i++; |
| 87 | + j++; |
| 88 | + continue; |
| 89 | + } |
| 90 | + |
| 91 | + i++; |
| 92 | + } |
| 93 | + |
| 94 | + return j === p.length; |
| 95 | +} |
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