LeetCode #1805 — EASY

Number of Different Integers in a String

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

The Problem

Problem Statement

You are given a string word that consists of digits and lowercase English letters.

You will replace every non-digit character with a space. For example, "a123bc34d8ef34" will become " 123 34 8 34". Notice that you are left with some integers that are separated by at least one space: "123", "34", "8", and "34".

Return the number of different integers after performing the replacement operations on word.

Two integers are considered different if their decimal representations without any leading zeros are different.

Example 1:

Input: word = "a123bc34d8ef34"
Output: 3
Explanation: The three different integers are "123", "34", and "8". Notice that "34" is only counted once.

Example 2:

Input: word = "leet1234code234"
Output: 2

Example 3:

Input: word = "a1b01c001"
Output: 1
Explanation: The three integers "1", "01", and "001" all represent the same integer because
the leading zeros are ignored when comparing their decimal values.

Constraints:

1 <= word.length <= 1000
word consists of digits and lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given a string word that consists of digits and lowercase English letters. You will replace every non-digit character with a space. For example, "a123bc34d8ef34" will become " 123 34 8 34". Notice that you are left with some integers that are separated by at least one space: "123", "34", "8", and "34". Return the number of different integers after performing the replacement operations on word. Two integers are considered different if their decimal representations without any leading zeros are different.

Baseline thinking

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

Pattern signal: Hash Map

Example 1

"a123bc34d8ef34"

Example 2

"leet1234code234"

Example 3

"a1b01c001"

Core Insight

What unlocks the optimal approach

Try to split the string so that each integer is in a different string.
Try to remove each integer's leading zeroes and compare the strings to find how many of them are unique.

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 #1805: Number of Different Integers in a String
class Solution {
    public int numDifferentIntegers(String word) {
        Set<String> s = new HashSet<>();
        int n = word.length();
        for (int i = 0; i < n; ++i) {
            if (Character.isDigit(word.charAt(i))) {
                while (i < n && word.charAt(i) == '0') {
                    ++i;
                }
                int j = i;
                while (j < n && Character.isDigit(word.charAt(j))) {
                    ++j;
                }
                s.add(word.substring(i, j));
                i = j;
            }
        }
        return s.size();
    }
}

// Accepted solution for LeetCode #1805: Number of Different Integers in a String
func numDifferentIntegers(word string) int {
	s := map[string]struct{}{}
	n := len(word)
	for i := 0; i < n; i++ {
		if word[i] >= '0' && word[i] <= '9' {
			for i < n && word[i] == '0' {
				i++
			}
			j := i
			for j < n && word[j] >= '0' && word[j] <= '9' {
				j++
			}
			s[word[i:j]] = struct{}{}
			i = j
		}
	}
	return len(s)
}

# Accepted solution for LeetCode #1805: Number of Different Integers in a String
class Solution:
    def numDifferentIntegers(self, word: str) -> int:
        s = set()
        i, n = 0, len(word)
        while i < n:
            if word[i].isdigit():
                while i < n and word[i] == '0':
                    i += 1
                j = i
                while j < n and word[j].isdigit():
                    j += 1
                s.add(word[i:j])
                i = j
            i += 1
        return len(s)

// Accepted solution for LeetCode #1805: Number of Different Integers in a String
use std::collections::HashSet;
impl Solution {
    pub fn num_different_integers(word: String) -> i32 {
        let s = word.as_bytes();
        let n = s.len();
        let mut set = HashSet::new();
        let mut i = 0;
        while i < n {
            if s[i] >= b'0' && s[i] <= b'9' {
                let mut j = i;
                while j < n && s[j] >= b'0' && s[j] <= b'9' {
                    j += 1;
                }
                while i < j - 1 && s[i] == b'0' {
                    i += 1;
                }
                set.insert(String::from_utf8(s[i..j].to_vec()).unwrap());
                i = j;
            } else {
                i += 1;
            }
        }
        set.len() as i32
    }
}

// Accepted solution for LeetCode #1805: Number of Different Integers in a String
function numDifferentIntegers(word: string): number {
    return new Set(
        word
            .replace(/\D+/g, ' ')
            .trim()
            .split(' ')
            .filter(v => v !== '')
            .map(v => v.replace(/^0+/g, '')),
    ).size;
}

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(n)

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.