From d87ab44d151150506c8311913da25c50ec673302 Mon Sep 17 00:00:00 2001
From: rain84 <rainy_sky@mail.ru>
Date: Tue, 14 Jan 2025 15:43:36 +0300
Subject: [PATCH] feat: add solutions to lc problem: No.2657

---
 .../README.md                                 | 56 +++++++++++++++++++
 .../README_EN.md                              | 56 +++++++++++++++++++
 .../Solution3.js                              | 17 ++++++
 .../Solution3.ts                              | 17 ++++++
 4 files changed, 146 insertions(+)
 create mode 100644 solution/2600-2699/2657.Find the Prefix Common Array of Two Arrays/Solution3.js
 create mode 100644 solution/2600-2699/2657.Find the Prefix Common Array of Two Arrays/Solution3.ts

diff --git a/solution/2600-2699/2657.Find the Prefix Common Array of Two Arrays/README.md b/solution/2600-2699/2657.Find the Prefix Common Array of Two Arrays/README.md
index e0bfe58e23e11..a639f233813c8 100644
--- a/solution/2600-2699/2657.Find the Prefix Common Array of Two Arrays/README.md	
+++ b/solution/2600-2699/2657.Find the Prefix Common Array of Two Arrays/README.md	
@@ -300,4 +300,60 @@ function findThePrefixCommonArray(A: number[], B: number[]): number[] {
 
 <!-- solution:end -->
 
+<!-- solution:start -->
+
+### Solution 3: Counting + Set
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of arrays $A$ and $B$.
+
+<!-- tabs:start -->
+
+#### TypeScript
+
+```ts
+function findThePrefixCommonArray(A: number[], B: number[]): number[] {
+    const n = A.length;
+    const res = Array(n).fill(0);
+    const [setA, setB] = [new Set<number>(), new Set<number>()];
+
+    for (let i = 0, c = 0; i < n; i++) {
+        if (setA.has(B[i])) c++;
+        if (setB.has(A[i])) c++;
+        if (A[i] === B[i]) c++;
+
+        setA.add(A[i]);
+        setB.add(B[i]);
+        res[i] = c;
+    }
+
+    return res;
+}
+```
+
+#### JavaScript
+
+```js
+function findThePrefixCommonArray(A, B) {
+    const n = A.length;
+    const res = Array(n).fill(0);
+    const [setA, setB] = [new Set(), new Set()];
+
+    for (let i = 0, c = 0; i < n; i++) {
+        if (setA.has(B[i])) c++;
+        if (setB.has(A[i])) c++;
+        if (A[i] === B[i]) c++;
+
+        setA.add(A[i]);
+        setB.add(B[i]);
+        res[i] = c;
+    }
+
+    return res;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
 <!-- problem:end -->
diff --git a/solution/2600-2699/2657.Find the Prefix Common Array of Two Arrays/README_EN.md b/solution/2600-2699/2657.Find the Prefix Common Array of Two Arrays/README_EN.md
index 12976a4d9baad..671be6f6f7ad6 100644
--- a/solution/2600-2699/2657.Find the Prefix Common Array of Two Arrays/README_EN.md	
+++ b/solution/2600-2699/2657.Find the Prefix Common Array of Two Arrays/README_EN.md	
@@ -300,4 +300,60 @@ function findThePrefixCommonArray(A: number[], B: number[]): number[] {
 
 <!-- solution:end -->
 
+<!-- solution:start -->
+
+### Solution 3: Counting + Set
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of arrays $A$ and $B$.
+
+<!-- tabs:start -->
+
+#### TypeScript
+
+```ts
+function findThePrefixCommonArray(A: number[], B: number[]): number[] {
+    const n = A.length;
+    const res = Array(n).fill(0);
+    const [setA, setB] = [new Set<number>(), new Set<number>()];
+
+    for (let i = 0, c = 0; i < n; i++) {
+        if (setA.has(B[i])) c++;
+        if (setB.has(A[i])) c++;
+        if (A[i] === B[i]) c++;
+
+        setA.add(A[i]);
+        setB.add(B[i]);
+        res[i] = c;
+    }
+
+    return res;
+}
+```
+
+#### JavaScript
+
+```js
+function findThePrefixCommonArray(A, B) {
+    const n = A.length;
+    const res = Array(n).fill(0);
+    const [setA, setB] = [new Set(), new Set()];
+
+    for (let i = 0, c = 0; i < n; i++) {
+        if (setA.has(B[i])) c++;
+        if (setB.has(A[i])) c++;
+        if (A[i] === B[i]) c++;
+
+        setA.add(A[i]);
+        setB.add(B[i]);
+        res[i] = c;
+    }
+
+    return res;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
 <!-- problem:end -->
diff --git a/solution/2600-2699/2657.Find the Prefix Common Array of Two Arrays/Solution3.js b/solution/2600-2699/2657.Find the Prefix Common Array of Two Arrays/Solution3.js
new file mode 100644
index 0000000000000..d8cf56614c4c5
--- /dev/null
+++ b/solution/2600-2699/2657.Find the Prefix Common Array of Two Arrays/Solution3.js	
@@ -0,0 +1,17 @@
+function findThePrefixCommonArray(A, B) {
+    const n = A.length;
+    const res = Array(n).fill(0);
+    const [setA, setB] = [new Set(), new Set()];
+
+    for (let i = 0, c = 0; i < n; i++) {
+        if (setA.has(B[i])) c++;
+        if (setB.has(A[i])) c++;
+        if (A[i] === B[i]) c++;
+
+        setA.add(A[i]);
+        setB.add(B[i]);
+        res[i] = c;
+    }
+
+    return res;
+}
diff --git a/solution/2600-2699/2657.Find the Prefix Common Array of Two Arrays/Solution3.ts b/solution/2600-2699/2657.Find the Prefix Common Array of Two Arrays/Solution3.ts
new file mode 100644
index 0000000000000..062f87b06adb6
--- /dev/null
+++ b/solution/2600-2699/2657.Find the Prefix Common Array of Two Arrays/Solution3.ts	
@@ -0,0 +1,17 @@
+function findThePrefixCommonArray(A: number[], B: number[]): number[] {
+    const n = A.length;
+    const res = Array(n).fill(0);
+    const [setA, setB] = [new Set<number>(), new Set<number>()];
+
+    for (let i = 0, c = 0; i < n; i++) {
+        if (setA.has(B[i])) c++;
+        if (setB.has(A[i])) c++;
+        if (A[i] === B[i]) c++;
+
+        setA.add(A[i]);
+        setB.add(B[i]);
+        res[i] = c;
+    }
+
+    return res;
+}