Number of Different Integers in a String

Number of Different Integers in a String - Problem

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.

Input & Output

Example 1 — Basic Case

$ Input: word = "a123bc34d8ef34"

› Output: 3

💡 Note: After replacing letters with spaces: " 123 34 8 34". The unique numbers are: "123", "34", "8". Note that "34" appears twice but counts as one unique number.

Example 2 — Leading Zeros

$ Input: word = "leet1234code234"

› Output: 2

💡 Note: After replacement: " 1234 234". The unique numbers are: "1234" and "234".

Example 3 — Leading Zeros Normalization

$ Input: word = "a1b01c001"

› Output: 1

💡 Note: After replacement: " 1 01 001". All three numbers normalize to "1" after removing leading zeros, so there's only 1 unique number.

Constraints

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

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is to extract numbers from the string and normalize them by removing leading zeros, then use a hash set for automatic deduplication. Best approach is the split and hash set method which cleanly separates string processing from uniqueness logic. Time: O(n), Space: O(k) where k is the number of unique integers.

Common Approaches

✓ Character-by-Character Parsing

⏱️ Time: O(n) Space: O(k)

Process each character individually, building numbers digit by digit. When encountering a non-digit, finalize the current number, normalize it by removing leading zeros, and add to a list. Finally count unique numbers.

Split and Hash Set

⏱️ Time: O(n) Space: O(n)

First replace all non-digit characters with spaces, then split the string by spaces to get individual number strings. Normalize each by converting to integer and back to string to remove leading zeros, then use a hash set to count unique numbers.

Character-by-Character Parsing — Algorithm Steps

Iterate through each character
Build numbers character by character
Normalize by removing leading zeros
Store in list and count unique values

Visualization

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Step-by-Step Walkthrough

Scan Characters

Process each character, building numbers

Normalize Numbers

Remove leading zeros from extracted numbers

Count Unique

Use set to count distinct normalized numbers

Code -

solution.c — C

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

#define MAX_NUMBERS 1000
#define MAX_NUM_LEN 100

int solution(char* word) {
    char numbers[MAX_NUMBERS][MAX_NUM_LEN];
    int count = 0;
    char currentNum[MAX_NUM_LEN] = "";
    int numLen = 0;
    
    for (int i = 0; word[i]; i++) {
        if (isdigit(word[i])) {
            currentNum[numLen++] = word[i];
        } else {
            if (numLen > 0) {
                currentNum[numLen] = '\0';
                // Remove leading zeros
                long long num = atoll(currentNum);
                sprintf(currentNum, "%lld", num);
                
                // Check if already exists
                int found = 0;
                for (int j = 0; j < count; j++) {
                    if (strcmp(numbers[j], currentNum) == 0) {
                        found = 1;
                        break;
                    }
                }
                if (!found) {
                    strcpy(numbers[count++], currentNum);
                }
                numLen = 0;
            }
        }
    }
    
    // Handle last number
    if (numLen > 0) {
        currentNum[numLen] = '\0';
        long long num = atoll(currentNum);
        sprintf(currentNum, "%lld", num);
        
        int found = 0;
        for (int j = 0; j < count; j++) {
            if (strcmp(numbers[j], currentNum) == 0) {
                found = 1;
                break;
            }
        }
        if (!found) {
            count++;
        }
    }
    
    return count;
}

int main() {
    char word[10000];
    fgets(word, sizeof(word), stdin);
    word[strcspn(word, "\n")] = 0;
    
    int result = solution(word);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through string of length n

✓ Linear Growth

Space Complexity

O(k)

Storage for k unique numbers extracted from string

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Character-by-Character Parsing — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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