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Omkarnath Parida
2020-10-04 01:06:25 +05:30
parent b8a519399a
commit a56758580d
1 changed files with 18 additions and 13 deletions
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Dynamic-Programming/LongestCommonSubsequence.js
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@ -1,31 +1,36 @@
/*
* 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.
* 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;
function longestCommonSubsequence (x, y, str1, str2, dp) {
if (x === -1 || y === -1) {
return 0
}
else {
if (dp[x][y] != 0) return dp[x][y];
if (dp[x][y] !== 0){
return dp[x][y]
}
else {
if (str1[x] == str2[y]) {
return dp[x][y] = 1 + longestCommonSubsequence(x - 1, y - 1, str1, str2, dp);
if (str1[x] === str2[y]) {
dp[x][y] = 1 + longestCommonSubsequence(x - 1, y - 1, str1, str2, dp);
return dp[x][y]
}
else {
return dp[x][y] = Math.max(longestCommonSubsequence(x - 1, y, str1, str2, dp), longestCommonSubsequence(x, y - 1, str1, str2, dp))
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"
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);
console.log(res)
}
main()