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optimised TC of longest increasing subsequence #1146

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Original file line number Diff line number Diff line change
@@ -1,53 +1,41 @@
/**
* Dynamic programming approach to find longest increasing subsequence.
* Complexity: O(n * n)
* Optimized approach to find longest increasing subsequence.
* Complexity: O(n * log n)
*
* @param {number[]} sequence
* @return {number}
*/
export default function dpLongestIncreasingSubsequence(sequence) {
// Create array with longest increasing substrings length and
// fill it with 1-s that would mean that each element of the sequence
// is itself a minimum increasing subsequence.
const lengthsArray = Array(sequence.length).fill(1);

let previousElementIndex = 0;
let currentElementIndex = 1;
if (sequence.length === 0) {
return 0;
}

while (currentElementIndex < sequence.length) {
if (sequence[previousElementIndex] < sequence[currentElementIndex]) {
// If current element is bigger then the previous one then
// current element is a part of increasing subsequence which
// length is by one bigger then the length of increasing subsequence
// for previous element.
const newLength = lengthsArray[previousElementIndex] + 1;
if (newLength > lengthsArray[currentElementIndex]) {
// Increase only if previous element would give us bigger subsequence length
// then we already have for current element.
lengthsArray[currentElementIndex] = newLength;
}
}
// This will store the smallest tail value for all increasing subsequences
// with length i+1 in tail[i].
const tails = [];

// Move previous element index right.
previousElementIndex += 1;
sequence.forEach((num) => {
let left = 0;
let right = tails.length;

// If previous element index equals to current element index then
// shift current element right and reset previous element index to zero.
if (previousElementIndex === currentElementIndex) {
currentElementIndex += 1;
previousElementIndex = 0;
// Binary search for the insertion point of the current element.
while (left < right) {
const mid = Math.floor((left + right) / 2);
if (tails[mid] < num) {
left = mid + 1;
} else {
right = mid;
}
}
}

// Find the biggest element in lengthsArray.
// This number is the biggest length of increasing subsequence.
let longestIncreasingLength = 0;

for (let i = 0; i < lengthsArray.length; i += 1) {
if (lengthsArray[i] > longestIncreasingLength) {
longestIncreasingLength = lengthsArray[i];
// If left is equal to the length of tails, it means we are adding a new subsequence.
if (left === tails.length) {
tails.push(num);
} else {
// Otherwise, we are updating an existing subsequence with a new tail value.
tails[left] = num;
}
}
});

return longestIncreasingLength;
return tails.length;
}