Long Pressed Name

Long Pressed Name - Problem

Your friend is typing his name into a keyboard. Sometimes, when typing a character c, the key might get long pressed, and the character will be typed 1 or more times.

You examine the typed characters of the keyboard. Return true if it is possible that it was your friend's name, with some characters (possibly none) being long pressed.

Input & Output

Example 1 — Basic Long Press

$ Input: name = "alex", typed = "aaleex"

› Output: true

💡 Note: The 'a' and 'e' were long pressed, resulting in extra characters, but the sequence matches the name

Example 2 — Wrong Character

$ Input: name = "saeed", typed = "ssaaedd"

› Output: false

💡 Note: The 'e' in typed was long pressed, but we're missing one 'e' that should appear after 'a'

Example 3 — Perfect Match

$ Input: name = "leelee", typed = "lleeelee"

› Output: true

💡 Note: All characters match with some long presses on 'l' and 'e'

Constraints

1 ≤ name.length ≤ 1000
1 ≤ typed.length ≤ 1000
name and typed consist of only lowercase English letters

Visualization

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Asked in

G Google 15 f Facebook 12

The key insight is to use two pointers to traverse both strings simultaneously, allowing for long-pressed characters. Best approach uses two pointers technique. Time: O(n + m), Space: O(1)

Common Approaches

✓ Brute Force

⏱️ Time: O(2^n) Space: O(n)

For each character in name, search through typed string to find matching sequences. This approach tries every possible way to match characters without optimization.

Two Pointers

⏱️ Time: O(n + m) Space: O(1)

Maintain two pointers, one for each string. Match characters one by one, allowing multiple occurrences in typed string for each character in name.

Brute Force — Algorithm Steps

Step 1: For each character in name, find all occurrences in typed
Step 2: Try all combinations to see if any valid matching exists

Visualization

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Step-by-Step Walkthrough

Start

Begin with first character of name

Branch

Try all possible matches in typed string

Backtrack

If no match found, backtrack and try different path

Code -

solution.c — C

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

bool backtrack(const char* name, const char* typed, int i, int j, int nameLen, int typedLen) {
    if (i == nameLen) {
        return j == typedLen;
    }
    if (j >= typedLen) {
        return false;
    }
    
    if (name[i] != typed[j]) {
        return false;
    }
    
    // Try consuming one or more of the same character in typed
    int k = j;
    while (k < typedLen && typed[k] == name[i]) {
        if (backtrack(name, typed, i + 1, k + 1, nameLen, typedLen)) {
            return true;
        }
        k++;
    }
    return false;
}

bool solution(char* name, char* typed) {
    return backtrack(name, typed, 0, 0, strlen(name), strlen(typed));
}

int main() {
    char name[1001], typed[1001];
    fgets(name, sizeof(name), stdin);
    fgets(typed, sizeof(typed), stdin);
    name[strcspn(name, "\n")] = 0;
    typed[strcspn(typed, "\n")] = 0;
    
    bool result = solution(name, typed);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2^n)

In worst case, we try all possible ways to match characters exponentially

⚠ Quadratic Growth

Space Complexity

O(n)

Recursion stack for backtracking through all possibilities

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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