64 lines
2.0 KiB
JavaScript
64 lines
2.0 KiB
JavaScript
"use strict";
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Object.defineProperty(exports, "__esModule", {
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value: true
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});
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exports.findRepeatedSubsequences = findRepeatedSubsequences;
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/**
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* Copyright (c) Microsoft Corporation.
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*
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* Licensed under the Apache License, Version 2.0 (the "License");
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* you may not use this file except in compliance with the License.
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* You may obtain a copy of the License at
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*
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* http://www.apache.org/licenses/LICENSE-2.0
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*
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* Unless required by applicable law or agreed to in writing, software
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* distributed under the License is distributed on an "AS IS" BASIS,
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* WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
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* See the License for the specific language governing permissions and
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* limitations under the License.
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*/
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function findRepeatedSubsequences(s) {
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const n = s.length;
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const result = [];
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let i = 0;
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const arraysEqual = (a1, a2) => {
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if (a1.length !== a2.length) return false;
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for (let j = 0; j < a1.length; j++) {
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if (a1[j] !== a2[j]) return false;
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}
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return true;
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};
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while (i < n) {
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let maxRepeatCount = 1;
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let maxRepeatSubstr = [s[i]]; // Initialize with the element at index i
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let maxRepeatLength = 1;
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// Try substrings of length from 1 to the remaining length of the array
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for (let p = 1; p <= n - i; p++) {
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const substr = s.slice(i, i + p); // Extract substring as array
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let k = 1;
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// Count how many times the substring repeats consecutively
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while (i + p * k <= n && arraysEqual(s.slice(i + p * (k - 1), i + p * k), substr)) k += 1;
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k -= 1; // Adjust k since it increments one extra time in the loop
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// Update the maximal repeating substring if necessary
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if (k > 1 && k * p > maxRepeatCount * maxRepeatLength) {
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maxRepeatCount = k;
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maxRepeatSubstr = substr;
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maxRepeatLength = p;
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}
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}
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// Record the substring and its count
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result.push({
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sequence: maxRepeatSubstr,
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count: maxRepeatCount
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});
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i += maxRepeatLength * maxRepeatCount; // Move index forward
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}
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return result;
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} |