feat: add solutions to lc problem: No.2657

doocs · Jan 14, 2025 · d87ab44 · d87ab44
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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
@@ -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
@@ -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
@@ -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
@@ -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;
+}