Find Occurrences of an Element in an Array

Find Occurrences of an Element in an Array - Problem

You are given an integer array nums, an integer array queries, and an integer x.

For each queries[i], you need to find the index of the queries[i]th occurrence of x in the nums array. If there are fewer than queries[i] occurrences of x, the answer should be -1 for that query.

Return an integer array answer containing the answers to all queries.

Input & Output

Example 1 — Basic Case

$ Input: nums = [1,3,1,1,2], queries = [1,3], x = 1

› Output: [0,3]

💡 Note: The 1st occurrence of 1 is at index 0. The 3rd occurrence of 1 is at index 3.

Example 2 — Query Not Found

$ Input: nums = [1,2,3], queries = [2], x = 1

› Output: [-1]

💡 Note: Element 1 appears only once at index 0, so the 2nd occurrence doesn't exist.

Example 3 — Multiple Queries

$ Input: nums = [1,2,1,2,1], queries = [1,2,3,4], x = 1

› Output: [0,2,4,-1]

💡 Note: 1st occurrence at index 0, 2nd at index 2, 3rd at index 4. 4th occurrence doesn't exist.

Constraints

1 ≤ nums.length ≤ 10⁵
1 ≤ queries.length ≤ 10⁴
1 ≤ queries[i] ≤ 10⁴
-10⁹ ≤ nums[i], x ≤ 10⁹

Visualization

Tap to expand

Understanding the Visualization

Input

Array nums, queries for occurrence positions, target element x

Process

Find positions of x, then lookup nth occurrences

Output

Array of indices for each query or -1 if not found

Key Takeaway

🎯 Key Insight: Preprocessing occurrence positions enables O(1) query responses

Asked in

G Google 15 a Amazon 12 f Meta 8

The key insight is to preprocess all positions where the target element appears, then answer queries in O(1) time. The optimal approach uses two-pass processing: first collect positions, then lookup. Time: O(n + m), Space: O(k) where k is occurrence count.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force - Search for Each Query	O(m × n)	O(1)	For each query, scan the array from start to find the nth occurrence
Preprocess Positions	O(n + m)	O(k)	First pass: collect all positions of x, then answer queries in O(1)

Brute Force - Search for Each Query — Algorithm Steps

For each query q
Scan nums from index 0
Count occurrences of x until we find the qth one
Return the index or -1 if not found

Visualization

Tap to expand

Step-by-Step Walkthrough

Query 1

Scan array to find 1st occurrence of x

Query 2

Scan array again to find 3rd occurrence of x

Result

Return array of indices

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int* solution(int* nums, int numsSize, int* queries, int queriesSize, int x, int* returnSize) {
    int* result = (int*)malloc(queriesSize * sizeof(int));
    *returnSize = queriesSize;
    
    for (int i = 0; i < queriesSize; i++) {
        int count = 0;
        int foundIndex = -1;
        
        for (int j = 0; j < numsSize; j++) {
            if (nums[j] == x) {
                count++;
                if (count == queries[i]) {
                    foundIndex = j;
                    break;
                }
            }
        }
        
        result[i] = foundIndex;
    }
    
    return result;
}

int main() {
    char line[10000];
    
    // Parse nums array
    fgets(line, sizeof(line), stdin);
    line[strcspn(line, "\n")] = 0;
    
    int nums[1000], numsSize = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    // Parse queries array
    fgets(line, sizeof(line), stdin);
    line[strcspn(line, "\n")] = 0;
    
    int queries[1000], queriesSize = 0;
    token = strtok(line + 1, ",]");
    while (token != NULL) {
        queries[queriesSize++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    // Parse x
    int x;
    scanf("%d", &x);
    
    int returnSize;
    int* result = solution(nums, numsSize, queries, queriesSize, x, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        printf("%d", result[i]);
        if (i < returnSize - 1) printf(",");
    }
    printf("]\n");
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(m × n)

For each of m queries, we scan up to n elements in nums array

✓ Linear Growth

Space Complexity

O(1)

Only using variables for counting, no extra data structures

✓ Linear Space

12.5K Views

Medium Frequency

~15 min Avg. Time

428 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Search for Each Query — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler