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Longest_Increasing_Subsequence/LIS.js

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+/**
+ * Dynamic programming approach to find longest increasing subsequence.
+ * Complexity: O(n * 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;
+
+  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;
+      }
+    }
+
+    // Move previous element index right.
+    previousElementIndex += 1;
+
+    // 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;
+    }
+  }
+
+  // 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];
+    }
+  }
+
+  return longestIncreasingLength;
+}