Solution #1092 - Mridul/Edited - 01/03/2025

Author: mrid88

DP/#1092 - Shortest Common Supersequence - Hard/Shortest Common Supersequence - Explanation.md

@@ -0,0 +1,53 @@
+class Solution {
+    public String shortestCommonSupersequence(String str1, String str2) {
+        int m = str1.length();
+        int n = str2.length();
+
+        // Create DP table
+        int[][] dp = new int[m + 1][n + 1];
+
+        // Fill the DP table
+        for (int i = 1; i <= m; i++) {
+            for (int j = 1; j <= n; j++) {
+                if (str1.charAt(i - 1) == str2.charAt(j - 1)) {
+                    dp[i][j] = dp[i - 1][j - 1] + 1; // If characters match
+                } else {
+                    dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]); // Take max from either previous row or column
+                }
+            }
+        }
+
+        // Build the shortest common supersequence
+        StringBuilder result = new StringBuilder();
+        int i = m, j = n;
+        while (i > 0 && j > 0) {
+            // If characters are the same, add to the result
+            if (str1.charAt(i - 1) == str2.charAt(j - 1)) {
+                result.append(str1.charAt(i - 1));
+                i--;
+                j--;
+            } else if (dp[i - 1][j] > dp[i][j - 1]) { // Move in the direction of the larger dp value
+                result.append(str1.charAt(i - 1));
+                i--;
+            } else {
+                result.append(str2.charAt(j - 1));
+                j--;
+            }
+        }
+
+        // If there are any remaining characters in str1
+        while (i > 0) {
+            result.append(str1.charAt(i - 1));
+            i--;
+        }
+
+        // If there are any remaining characters in str2
+        while (j > 0) {
+            result.append(str2.charAt(j - 1));
+            j--;
+        }
+
+        // Reverse the result before returning
+        return result.reverse().toString();
+    }
+}