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