Unique Number of Occurrences - Practice Coding Problems

Unique Number of Occurrences - Problem

Given an array of integers arr, return true if the number of occurrences of each value in the array is unique or false otherwise.

In other words: No two different values should have the same frequency count.

Input & Output

Example 1 — All Unique Frequencies

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

› Output: true

💡 Note: Element 1 appears 3 times, element 2 appears 2 times, element 3 appears 1 time. Frequencies [3,2,1] are all unique.

Example 2 — Duplicate Frequencies

$ Input: arr = [1,2]

› Output: false

💡 Note: Element 1 appears 1 time, element 2 appears 1 time. Both have frequency 1, so frequencies [1,1] are not unique.

Example 3 — Single Element

$ Input: arr = [5]

› Output: true

💡 Note: Only one element with frequency 1. Since there's only one frequency count, it's trivially unique.

Constraints

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

Visualization

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Understanding the Visualization

Input Array

Array of integers with repeated elements

Count Frequencies

Use hash map to count each element's occurrences

Check Uniqueness

Verify all frequency counts are different

Key Takeaway

🎯 Key Insight: Use a hash map to count frequencies, then check if all counts are unique using a set

Asked in

a Amazon 15 M Microsoft 12

The key insight is to count frequency of each element using a hash map, then check if all frequency counts are unique using a set. Best approach is the hash map with set method. Time: O(n), Space: O(n)

Common Approaches

Approach	Time	Space	Notes
✓ Frequency Count with Hash Map	O(n)	O(n)	Use hash map to count frequencies, then check if all counts are unique
Single Pass Optimization	O(n)	O(n)	Count frequencies and check uniqueness in a single optimized pass
Brute Force - Count and Compare All Pairs	O(n³)	O(1)	Count frequency of each element, then check all pairs of counts for duplicates

Frequency Count with Hash Map — Algorithm Steps

Build frequency map of all elements
Collect all frequency counts
Use set to check if all counts are unique

Visualization

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Step-by-Step Walkthrough

Build Frequency Map

Single pass to count each element

Extract Counts

Get all frequency values

Check Uniqueness

Use set to verify all counts are different

Code -

solution.c — C

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

#define MAX_SIZE 1001

bool solution(int* arr, int n) {
    int freqMap[MAX_SIZE] = {0};
    int frequencies[MAX_SIZE];
    int freqCount = 0;
    
    // Count frequency of each element (assuming values are small)
    for (int i = 0; i < n; i++) {
        if (arr[i] >= 0 && arr[i] < MAX_SIZE) {
            freqMap[arr[i]]++;
        }
    }
    
    // Collect frequencies
    for (int i = 0; i < MAX_SIZE; i++) {
        if (freqMap[i] > 0) {
            frequencies[freqCount++] = freqMap[i];
        }
    }
    
    // Check if all frequencies are unique
    for (int i = 0; i < freqCount; i++) {
        for (int j = i + 1; j < freqCount; j++) {
            if (frequencies[i] == frequencies[j]) {
                return false;
            }
        }
    }
    
    return true;
}

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

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass to build frequency map + single pass through frequencies

✓ Linear Growth

Space Complexity

O(n)

Hash map stores up to n elements, set stores up to n frequencies

⚡ Linearithmic Space

178.0K Views

Medium Frequency

~15 min Avg. Time

2.8K Likes

Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Frequency Count with Hash Map — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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