O(N*M)

Author: skjha1

src/Algorithms/Dynamic Programming/23 Print Shortest Common Supersequence.cpp

@@ -0,0 +1,58 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+string SCS(string X, string Y, int n, int m) {
+	int dp[n + 1][m + 1]; // DP - matrix
+
+	// base case of recursion --> for initialization of dp - matrix
+	for (int i = 0; i <= n; i++)
+		for (int j = 0; j <= m; j++)
+			if (i == 0 || j == 0)
+				dp[i][j] = 0;
+
+	for (int i = 1; i <= n; i++)
+		for (int j = 1; j <= m; j++)
+			if (X[i - 1] == Y[j - 1]) // when last character is same
+				dp[i][j] = 1 + dp[i - 1][j - 1];
+			else // when last character is not same -> pick max
+				dp[i][j] = max(dp[i][j - 1], dp[i - 1][j]);
+
+	int i = n, j = m;
+	string scs = "";
+	while (i > 0 && j > 0) {
+		if (X[i - 1] == Y[j - 1]) { // when char are eqaul from table obs 
+			scs += X[i - 1]; // take only once 
+			i--, j--; // and decrement both
+		}
+		else if (dp[i][j - 1] > dp[i - 1][j]) {
+			scs += Y[j - 1]; // in this we will take the charecter to string 
+			j--;
+		}
+		else {
+			scs += X[i - 1]; // taking the charecter to string 
+			i--;
+		}
+	}
+
+	while (i > 0) {
+		scs += X[i - 1]; // take left chareter from str1
+		i--;
+	}
+
+	while (j > 0) {
+		scs += Y[j - 1]; // take left chareter from str1
+		j--;
+	}
+
+	reverse(scs.begin(), scs.end());
+
+	return scs;
+}
+
+int main() {
+	string X, Y; cin >> X >> Y;
+	int n = X.length(), m = Y.length();
+
+	cout << SCS(X, Y, n, m) << endl;
+	return 0;
+}