DP - DeleteOperationsForTwoStrings

ankushk1729 · ankushk1729 · commit 09a6fbcfce4f · 2022-10-07T23:18:34.000+05:30
diff --git a/DynamicProgramming/DeleteOperationsForTwoStrings.js b/DynamicProgramming/DeleteOperationsForTwoStrings.js
@@ -0,0 +1,48 @@
+/*
+    Input: word1 = "sea", word2 = "eat"
+    Output: 2
+    Explanation: You need one step to make "sea" to "ea" and another step to make "eat" to "ea".
+*/
+
+
+function min(i, j){
+	return i > j ? j : i;
+}
+
+function max(i, j){
+	return i > j ? i : j;
+}
+
+function solve(w1, w2, dp, i, j){
+    // If we have traversed both the strings till the end at the same time,
+    // that means there are no extra characters, so return 0.
+    // (Is being handled by the below base case but just added to minimize a max 
+    // operation on some large string sizes)
+    if(i == w1.length && j == w2.length) return 0;
+    
+    // If one of the strings is traversed completely before the other, 
+    // that means there are some extra characters in the other string, 
+    // so we add the remaining length to our steps count.
+    if(i == w1.length || j == w2.length) return max(w1.length-i, w2.length-j);
+    
+    
+    // return the cached value.
+    if(dp[i][j] != -1) return dp[i][j];
+    
+    // if both characters are same that means no need to remove those characters
+    // from both the strings, so just increment the pointer for both strings.
+    if(w1[i] == w2[j]) return solve(w1, w2, dp, i + 1, j + 1);
+    
+    // otherwise we add one to the step count and solve for minimum steps
+    // from the options of removing from string w1 or w2.
+    return dp[i][j] = 1 + min(solve(w1, w2, dp, i + 1, j), solve(w1, w2, dp, i, j + 1));
+}
+function minDistance(word1, word2) {
+    let dp = Array(word1.length + 1).fill().map(() => Array(word2.length+1).fill(-1))
+    return solve(word1, word2, dp, 0, 0);
+}
+
+
+let s1 = "sea";
+let s2 = "eat";
+console.log(minDistance(s1, s2));
