Longest Common Subsequence - Amazon Top Interview Questions
Problem Statement :
Given two strings a and b, return the length of their longest common subsequence. Constraints n ≤ 1,000 where n is the length of a m ≤ 1,000 where m is the length of b Example 1 Input a = "abcvc" b = "bv" Output 2 Explanation bv is the longest common subsequence. Example 2 Input a = "abc" b = "abc" Output 3 Example 3 Input a = "abc" b = "def" Output 0 Example 4 Input a = "binarysearch" b = "searchbinary" Output 6
Solution :
Solution in C++ :
int helper(const string a, const string b, int i, int j) {
if (i < 0 || j < 0) return 0;
if (a[i] == b[j]) return 1 + helper(a, b, i - 1, j - 1);
return max(helper(a, b, i - 1, j), helper(a, b, i, j - 1));
}
int solve(string a, string b) {
int m = a.length(), n = b.length();
vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
if (a[i - 1] == b[j - 1])
dp[i][j] = dp[i - 1][j - 1] + 1;
else {
dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
}
}
}
return dp[m][n];
}
Solution in Python :
class Solution:
def solve(self, a, b):
# initializing a dp array of size
n = len(a)
m = len(b)
memo = [[0 for i in range(m + 1)] for i in range(n + 1)]
for i in range(n - 1, -1, -1):
for j in range(m - 1, -1, -1):
if a[i] == b[j]:
memo[i][j] = memo[i + 1][j + 1] + 1
else:
memo[i][j] = max(memo[i + 1][j], memo[i][j + 1])
return memo[0][0]
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