Counting Elements

Counting Elements - Problem

Given an integer array arr, count how many elements x there are, such that x + 1 is also in arr.

If there are duplicates in arr, count them separately.

Example: If arr = [1, 2, 3], then we have element 1 (and 1 + 1 = 2 exists), and element 2 (and 2 + 1 = 3 exists). So the answer is 2.

Input & Output

Example 1 — Basic Case

$ Input: arr = [1,2,3]

› Output: 2

💡 Note: Element 1 has successor 2 in array, element 2 has successor 3 in array. Element 3 has no successor 4. So count = 2.

Example 2 — With Duplicates

$ Input: arr = [1,1,3,3,5,5,7,7]

› Output: 0

💡 Note: No element x has x+1 in the array. 1→2 (missing), 3→4 (missing), 5→6 (missing), 7→8 (missing). Count = 0.

Example 3 — Counting Duplicates

$ Input: arr = [1,3,2,3,5,0]

› Output: 3

💡 Note: Element 0 has successor 1, element 1 has successor 2, element 2 has successor 3. Both instances of 3 don't have successor 4. Count = 3.

Constraints

1 ≤ arr.length ≤ 1000
0 ≤ arr[i] ≤ 1000

Visualization

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Asked in

F Facebook 15 A Amazon 12

The key insight is to use a hash set for O(1) successor lookups instead of linear search. Build a set of all elements, then count how many elements have their x+1 present. Best approach is hash set with Time: O(n), Space: O(n).

Common Approaches

✓ Brute Force

⏱️ Time: O(n²) Space: O(1)

For every element x in the array, iterate through the entire array again to check if x+1 exists. Count all such elements where the successor is found.

Hash Set Lookup

⏱️ Time: O(n) Space: O(n)

First pass creates a hash set of all elements for O(1) lookup. Second pass iterates through the array and for each element x, checks if x+1 exists in the set in constant time.

Brute Force — Algorithm Steps

Iterate through each element x in the array
For each x, scan the entire array to find x+1
If x+1 is found, increment counter

Visualization

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Step-by-Step Walkthrough

Select Element

Pick current element x from array

Linear Search

Scan entire array looking for x+1

Count Match

If x+1 found, increment counter

Code -

solution.c — C

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

int solution(int* arr, int size) {
    int count = 0;
    for (int i = 0; i < size; i++) {
        int x = arr[i];
        // Check if x+1 exists in the array
        for (int j = 0; j < size; j++) {
            if (arr[j] == x + 1) {
                count++;
                break;
            }
        }
    }
    return count;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int arr[1000];
    int size = 0;
    
    char* token = strtok(line, "[],\n");
    while (token != NULL) {
        if (strlen(token) > 0 && token[0] != ' ') {
            arr[size++] = atoi(token);
        }
        token = strtok(NULL, "[],\n");
    }
    
    printf("%d\n", solution(arr, size));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops - for each of n elements, we scan n elements to find successor

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a counter variable, no extra data structures

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

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Code -

Time & Space Complexity

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