Add longest increasing subsequence.

Author: trekhleb

README.md

@@ -40,6 +40,7 @@
   * [Combinations](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/sets/combinations) (with and without repetitions)
   * [Fisher–Yates Shuffle](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/sets/fisher-yates) - random permutation of a finite sequence
   * [Longest Common Subsequence](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/sets/longest-common-subsequnce) (LCS) 
+  * [Longest Increasing subsequence](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/sets/longest-increasing-subsequence)
 * **String**
   * [Levenshtein Distance](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/string/levenshtein-distance) - minimum edit distance between two sequences
   * [Hamming Distance](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/string/hamming-distance) - number of positions at which the symbols are different
@@ -92,8 +93,7 @@
   * [Levenshtein Distance](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/string/levenshtein-distance) - minimum edit distance between two sequences
   * [Longest Common Subsequence](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/sets/longest-common-subsequnce) (LCS)
   * [Longest Common Substring](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/string/longest-common-substring)
-  * Increasing subsequence
-  * Longest Increasing subsequence
+  * [Longest Increasing subsequence](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/sets/longest-increasing-subsequence)
   * Shortest common supersequence
   * Knapsack problem
   * Maximum subarray

src/algorithms/sets/longest-increasing-subsequence/README.md

@@ -0,0 +1,46 @@
+# Longest Increasing Subsequence
+
+The longest increasing subsequence problem is to find a subsequence of a 
+given sequence in which the subsequence's elements are in sorted order, 
+lowest to highest, and in which the subsequence is as long as possible. 
+This subsequence is not necessarily contiguous, or unique. 
+
+## Complexity
+
+The longest increasing subsequence problem is solvable in 
+time `O(n log n)`, where `n` denotes the length of the input sequence.
+
+Dynamic programming approach has complexity `O(n * n)`.
+
+## Example
+
+In the first 16 terms of the binary Van der Corput sequence
+
+```
+0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15
+```
+
+a longest increasing subsequence is
+
+```
+0, 2, 6, 9, 11, 15.
+```
+
+This subsequence has length six; 
+the input sequence has no seven-member increasing subsequences. 
+The longest increasing subsequence in this example is not unique: for 
+instance,
+
+```
+0, 4, 6, 9, 11, 15 or
+0, 2, 6, 9, 13, 15 or
+0, 4, 6, 9, 13, 15
+```
+
+are other increasing subsequences of equal length in the same 
+input sequence.
+
+## References
+
+- [Wikipedia](https://en.wikipedia.org/wiki/Longest_increasing_subsequence)
+- [Dynamic Programming Approach on YouTube](https://www.youtube.com/watch?v=CE2b_-XfVDk)

src/algorithms/sets/longest-increasing-subsequence/__test__/dpLongestIncreasingSubsequence.test.js

@@ -0,0 +1,36 @@
+import dpLongestIncreasingSubsequence from '../dpLongestIncreasingSubsequence';
+
+describe('dpLongestIncreasingSubsequence', () => {
+  it('should find longest increasing subsequence length', () => {
+    // Should be:
+    // 9 or
+    // 8 or
+    // 7 or
+    // 6 or
+    // ...
+    expect(dpLongestIncreasingSubsequence([
+      9, 8, 7, 6, 5, 4, 3, 2, 1, 0,
+    ])).toBe(1);
+
+    // Should be:
+    // 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
+    expect(dpLongestIncreasingSubsequence([
+      0, 1, 2, 3, 4, 5, 6, 7, 8, 9,
+    ])).toBe(10);
+
+    // Should be:
+    // -1, 0, 2, 3
+    expect(dpLongestIncreasingSubsequence([
+      3, 4, -1, 0, 6, 2, 3,
+    ])).toBe(4);
+
+    // Should be:
+    // 0, 2, 6, 9, 11, 15 or
+    // 0, 4, 6, 9, 11, 15 or
+    // 0, 2, 6, 9, 13, 15 or
+    // 0, 4, 6, 9, 13, 15
+    expect(dpLongestIncreasingSubsequence([
+      0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15,
+    ])).toBe(6);
+  });
+});

src/algorithms/sets/longest-increasing-subsequence/dpLongestIncreasingSubsequence.js

@@ -0,0 +1,53 @@
+/**
+ * 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;
+}