Lis

Dynamic-Programming/LongestCommonSubsequence.js

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+/*
+    * Given two sequences, find the length of longest subsequence present in both of them.
+    * A subsequence is a sequence that appears in the same relative order, but not necessarily contiguous.
+    * For example, “abc”, “abg”, “bdf”, “aeg”, ‘”acefg”, .. etc are subsequences of “abcdefg”
+*/
+
+function longestCommonSubsequence (x, y, str1, str2, dp) {
+  if (x === -1 || y === -1) {
+    return 0
+  } else {
+    if (dp[x][y] !== 0) {
+      return dp[x][y]
+    } else {
+      if (str1[x] === str2[y]) {
+        dp[x][y] = 1 + longestCommonSubsequence(x - 1, y - 1, str1, str2, dp)
+        return dp[x][y]
+      } else {
+        dp[x][y] = Math.max(longestCommonSubsequence(x - 1, y, str1, str2, dp), longestCommonSubsequence(x, y - 1, str1, str2, dp))
+        return dp[x][y]
+      }
+    }
+  }
+}
+
+function main () {
+  const str1 = 'ABCDGH'
+  const str2 = 'AEDFHR'
+  const dp = new Array(str1.length + 1).fill(0).map(x => new Array(str2.length + 1).fill(0))
+  const res = longestCommonSubsequence(str1.length - 1, str2.length - 1, str1, str2, dp)
+  console.log(res)
+}
+
+main()

Dynamic-Programming/LongestIncreasingSubsequence.js

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+/**
+ * A Dynamic Programming based solution for calculating Longest Increasing Subsequence
+ * https://en.wikipedia.org/wiki/Longest_increasing_subsequence
+ */
+
+function main () {
+  const x = [0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15]
+  const length = x.length
+  const dp = Array(length).fill(1)
+
+  let res = 1
+
+  for (let i = 0; i < length; i++) {
+    for (let j = 0; j < i; j++) {
+      if (x[i] > x[j]) {
+        dp[i] = Math.max(dp[i], 1 + dp[j])
+        if (dp[i] > res) {
+          res = dp[i]
+        }
+      }
+    }
+  }
+
+  console.log('Length of Longest Increasing Subsequence is:', res)
+}
+
+main()