Improved task 30

README.md

@@ -1527,7 +1527,7 @@ implementation 'com.github.javadev:leetcode-in-kotlin:1.34'
 |-|-|-|-|-|-
 | 0209 |[Minimum Size Subarray Sum](src/main/kotlin/g0201_0300/s0209_minimum_size_subarray_sum/Solution.kt)| Medium | Array, Binary_Search, Prefix_Sum, Sliding_Window | 315 | 96.73
 | 0003 |[Longest Substring Without Repeating Characters](src/main/kotlin/g0001_0100/s0003_longest_substring_without_repeating_characters/Solution.kt)| Medium | Top_100_Liked_Questions, Top_Interview_Questions, String, Hash_Table, Sliding_Window, Big_O_Time_O(n)_Space_O(1), AI_can_be_used_to_solve_the_task | 201 | 87.28
-| 0030 |[Substring with Concatenation of All Words](src/main/kotlin/g0001_0100/s0030_substring_with_concatenation_of_all_words/Solution.kt)| Hard | String, Hash_Table, Sliding_Window | 182 | 100.00
+| 0030 |[Substring with Concatenation of All Words](src/main/kotlin/g0001_0100/s0030_substring_with_concatenation_of_all_words/Solution.kt)| Hard | String, Hash_Table, Sliding_Window | 14 | 98.62
 | 0076 |[Minimum Window Substring](src/main/kotlin/g0001_0100/s0076_minimum_window_substring/Solution.kt)| Hard | Top_100_Liked_Questions, Top_Interview_Questions, String, Hash_Table, Sliding_Window, Big_O_Time_O(s.length())_Space_O(1) | 191 | 96.38
 
 #### Top Interview 150 Matrix

src/main/kotlin/g0001_0100/s0030_substring_with_concatenation_of_all_words/Solution.kt

@@ -1,47 +1,44 @@
 package g0001_0100.s0030_substring_with_concatenation_of_all_words
 
 // #Hard #String #Hash_Table #Sliding_Window #Top_Interview_150_Sliding_Window
-// #2023_07_03_Time_182_ms_(100.00%)_Space_37.9_MB_(100.00%)
+// #2025_03_04_Time_14_ms_(98.62%)_Space_39.70_MB_(91.72%)
 
 class Solution {
     fun findSubstring(s: String, words: Array<String>): List<Int> {
-        val indices: MutableList<Int> = ArrayList()
-        if (words.size == 0) {
-            return indices
+        val ans: MutableList<Int> = ArrayList<Int>()
+        val n1 = words[0].length
+        val n2 = s.length
+        val map1: MutableMap<String, Int> = HashMap<String, Int>()
+        for (ch in words) {
+            map1.put(ch, map1.getOrDefault(ch, 0) + 1)
         }
-        // Put each word into a HashMap and calculate word frequency
-        val wordMap: MutableMap<String, Int> = HashMap()
-        for (word in words) {
-            wordMap[word] = wordMap.getOrDefault(word, 0) + 1
-        }
-        val wordLength = words[0].length
-        val window = words.size * wordLength
-        for (i in 0 until wordLength) {
-            // move a word's length each time
+        for (i in 0..<n1) {
+            var left = i
             var j = i
-            while (j + window <= s.length) {
-                // get the subStr
-                val subStr = s.substring(j, j + window)
-                val map: MutableMap<String, Int> = HashMap()
-                // start from the last word
-                for (k in words.indices.reversed()) {
-                    // get the word from subStr
-                    val word = subStr.substring(k * wordLength, (k + 1) * wordLength)
-                    val count = map.getOrDefault(word, 0) + 1
-                    // if the num of the word is greater than wordMap's, move (k * wordLength) and
-                    // break
-                    if (count > wordMap.getOrDefault(word, 0)) {
-                        j = j + k * wordLength
-                        break
-                    } else if (k == 0) {
-                        indices.add(j)
-                    } else {
-                        map[word] = count
+            var c = 0
+            val map2: MutableMap<String, Int> = HashMap<String, Int>()
+            while (j + n1 <= n2) {
+                val word1 = s.substring(j, j + n1)
+                j += n1
+                if (map1.containsKey(word1)) {
+                    map2.put(word1, map2.getOrDefault(word1, 0) + 1)
+                    c++
+                    while (map2[word1]!! > map1[word1]!!) {
+                        val word2 = s.substring(left, left + n1)
+                        map2.put(word2, map2[word2]!! - 1)
+                        left += n1
+                        c--
+                    }
+                    if (c == words.size) {
+                        ans.add(left)
                     }
+                } else {
+                    map2.clear()
+                    c = 0
+                    left = j
                 }
-                j = j + wordLength
             }
         }
-        return indices
+        return ans
     }
 }