Update 1143-longest-common-subsequence.kt

drxlx · drxlx · commit b84c9276797e · 2024-10-23T09:13:24.000+03:00
diff --git a/kotlin/1143-longest-common-subsequence.kt b/kotlin/1143-longest-common-subsequence.kt
@@ -20,79 +20,4 @@ class Solution {
 		}
 		return dp[M][N]
 	}
-}
-
-/*
- * Different solutions
- */
-
- // Recursion + Memoization, Time Complexity of O(n * m) and space complexity of O(n * m)
-class Solution {
-    fun longestCommonSubsequence(t1: String, t2: String): Int {
-        val n = t1.length
-        val m = t2.length
-        val dp = Array (n) { IntArray (m) { -1 } }
-
-        fun dfs(i: Int, j: Int): Int {
-            if (i == n || j == m) return 0
-            if (dp[i][j] != -1) return dp[i][j]
-
-            if (t1[i] == t2[j])
-                dp[i][j] = 1 + dfs(i + 1, j + 1)
-            else
-                dp[i][j] = maxOf(dfs(i + 1, j), dfs(i, j + 1))
-
-            return dp[i][j]
-        }
-
-        return dfs(0, 0)
-    }
-}
-
-// Top down DP, Time Complexity of O(n * m) and space complexity of O(n * m)
-class Solution {
-    fun longestCommonSubsequence(t1: String, t2: String): Int {
-        val n = t1.length
-        val m = t2.length
-        val dp = Array (n + 1) { IntArray (m + 1) }
-
-        for (i in n - 1 downTo 0) {
-            for (j in m - 1 downTo 0) {
-                if (t1[i] == t2[j])
-                    dp[i][j] = 1 + dp[i + 1][j + 1]
-                else
-                    dp[i][j] = maxOf(dp[i + 1][j], dp[i][j + 1])
-            }
-        }
-
-        return dp[0][0]
-    }
-}
-
-// Optimized DP (Works both for both Top-down and Bottom-up, but here we use bottom-up approach)
-// Time Complexity of O(n * m) and space complexity of O(maxOf(n, m))
-class Solution {
-    fun longestCommonSubsequence(t1: String, t2: String): Int {
-        val m = t1.length
-        val n = t2.length
-        if (m < n) return longestCommonSubsequence(t2, t1)
-
-        var dp = IntArray (n + 1)
-
-        for (i in m downTo 0) {
-            var newDp = IntArray (n + 1)
-            for (j in n downTo 0) {
-                if (i == m || j == n) {
-                    newDp[j] = 0
-                } else if (t1[i] == t2[j]) {
-                    newDp[j] = 1 + dp[j + 1]
-                } else {
-                    newDp[j] = maxOf(dp[j], newDp[j + 1])
-                }
-            }
-            dp = newDp
-        }
-
-        return dp[0]
-    }
-}
+}
