LeetCode #2283 — EASY

Check if Number Has Equal Digit Count and Digit Value

Build confidence with an intuition-first walkthrough focused on hash map fundamentals.

The Problem

Problem Statement

You are given a 0-indexed string num of length n consisting of digits.

Return true if for every index i in the range 0 <= i < n, the digit i occurs num[i] times in num, otherwise return false.

Example 1:

Input: num = "1210"
Output: true
Explanation:
num[0] = '1'. The digit 0 occurs once in num.
num[1] = '2'. The digit 1 occurs twice in num.
num[2] = '1'. The digit 2 occurs once in num.
num[3] = '0'. The digit 3 occurs zero times in num.
The condition holds true for every index in "1210", so return true.

Example 2:

Input: num = "030"
Output: false
Explanation:
num[0] = '0'. The digit 0 should occur zero times, but actually occurs twice in num.
num[1] = '3'. The digit 1 should occur three times, but actually occurs zero times in num.
num[2] = '0'. The digit 2 occurs zero times in num.
The indices 0 and 1 both violate the condition, so return false.

Constraints:

n == num.length
1 <= n <= 10
num consists of digits.

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed string num of length n consisting of digits. Return true if for every index i in the range 0 <= i < n, the digit i occurs num[i] times in num, otherwise return false.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map

Example 1

"1210"

Example 2

"030"

Core Insight

What unlocks the optimal approach

Count the frequency of each digit in num.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2283: Check if Number Has Equal Digit Count and Digit Value
class Solution {
    public boolean digitCount(String num) {
        int[] cnt = new int[10];
        int n = num.length();
        for (int i = 0; i < n; ++i) {
            ++cnt[num.charAt(i) - '0'];
        }
        for (int i = 0; i < n; ++i) {
            if (num.charAt(i) - '0' != cnt[i]) {
                return false;
            }
        }
        return true;
    }
}

// Accepted solution for LeetCode #2283: Check if Number Has Equal Digit Count and Digit Value
func digitCount(num string) bool {
	cnt := [10]int{}
	for _, c := range num {
		cnt[c-'0']++
	}
	for i, c := range num {
		if int(c-'0') != cnt[i] {
			return false
		}
	}
	return true
}

# Accepted solution for LeetCode #2283: Check if Number Has Equal Digit Count and Digit Value
class Solution:
    def digitCount(self, num: str) -> bool:
        cnt = Counter(int(x) for x in num)
        return all(cnt[i] == int(x) for i, x in enumerate(num))

// Accepted solution for LeetCode #2283: Check if Number Has Equal Digit Count and Digit Value
impl Solution {
    pub fn digit_count(num: String) -> bool {
        let mut cnt = vec![0; 10];
        for c in num.chars() {
            let x = c.to_digit(10).unwrap() as usize;
            cnt[x] += 1;
        }
        for (i, c) in num.chars().enumerate() {
            let x = c.to_digit(10).unwrap() as usize;
            if cnt[i] != x {
                return false;
            }
        }
        true
    }
}

// Accepted solution for LeetCode #2283: Check if Number Has Equal Digit Count and Digit Value
function digitCount(num: string): boolean {
    const cnt: number[] = Array(10).fill(0);
    for (const c of num) {
        ++cnt[+c];
    }
    for (let i = 0; i < num.length; ++i) {
        if (cnt[i] !== +num[i]) {
            return false;
        }
    }
    return true;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(|\Sigma|)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan the rest of the array looking for a match. Two nested loops give n × (n−1)/2 comparisons = O(n²). No extra space since we only use loop indices.

HASH MAP

O(n) time

O(n) space

One pass through the input, performing O(1) hash map lookups and insertions at each step. The hash map may store up to n entries in the worst case. This is the classic space-for-time tradeoff: O(n) extra memory eliminates an inner loop.

Shortcut: Need to check “have I seen X before?” → hash map → O(n) time, O(n) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.