Minimum Number of Pushes to Type Word I - Problem

You are given a string word containing distinct lowercase English letters.

Telephone keypads have keys mapped with distinct collections of lowercase English letters, which can be used to form words by pushing them. For example, the key 2 is mapped with ["a","b","c"], we need to push the key one time to type "a", two times to type "b", and three times to type "c".

It is allowed to remap the keys numbered 2 to 9 to distinct collections of letters. The keys can be remapped to any amount of letters, but each letter must be mapped to exactly one key.

You need to find the minimum number of times the keys will be pushed to type the string word.

Return the minimum number of pushes needed to type word after remapping the keys.

Input & Output

Example 1 — Basic Case

$ Input: word = "abcde"

› Output: 5

💡 Note: Each character appears once. Assign each to position 1 on different keys: a(1×1) + b(1×1) + c(1×1) + d(1×1) + e(1×1) = 5

Example 2 — Frequency Matters

$ Input: word = "aabbbc"

› Output: 7

💡 Note: c appears 3 times, b appears 2 times, a appears 2 times. Assign c, b, a to position 1 on keys 2, 3, 4: c(3×1) + b(2×1) + a(2×1) = 7

Example 3 — Single Character

$ Input: word = "aaa"

› Output: 3

💡 Note: Only character 'a' appears 3 times. Assign to position 1: a(3×1) = 3

Constraints

1 ≤ word.length ≤ 10⁵
word consists of distinct lowercase English letters

Visualization

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Asked in

G Google 25 M Microsoft 20 a Amazon 15

The key insight is to use a greedy approach: assign the most frequently occurring characters to keypad positions that require the fewest presses. Best approach counts character frequencies and assigns them optimally. Time: O(n + k log k), Space: O(k).

Common Approaches

✓ Brute Force - Character Counting

⏱️ Time: O(n) Space: O(1)

Count how many times each character appears in the word, then assign characters to keypads arbitrarily. Calculate total presses needed based on position of each character on its assigned key.

Greedy Optimal Assignment

⏱️ Time: O(n + k log k) Space: O(k)

Count character frequencies, then greedily assign the most frequent characters to positions that require the fewest key presses (position 1 on each key, then position 2, etc.).

Brute Force - Character Counting — Algorithm Steps

Step 1: Count frequency of each character
Step 2: Assign characters to keys 2-9 arbitrarily
Step 3: Calculate total presses based on position

Visualization

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Step-by-Step Walkthrough

Count Characters

Count frequency of each character in word

Assign Positions

Assign characters to keypad positions arbitrarily

Calculate

Sum up total presses needed

Code -

solution.c — C

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

typedef struct {
    char ch;
    int freq;
} CharFreq;

int compare(const void* a, const void* b) {
    CharFreq* cf1 = (CharFreq*)a;
    CharFreq* cf2 = (CharFreq*)b;
    return cf2->freq - cf1->freq;  // Descending order
}

int solution(char* word) {
    // Count frequency of each character
    int freq[26] = {0};
    int len = strlen(word);
    
    for (int i = 0; i < len; i++) {
        freq[word[i] - 'a']++;
    }
    
    // Create array of character frequencies
    CharFreq charFreqs[26];
    int count = 0;
    
    for (int i = 0; i < 26; i++) {
        if (freq[i] > 0) {
            charFreqs[count].ch = 'a' + i;
            charFreqs[count].freq = freq[i];
            count++;
        }
    }
    
    // Sort by frequency (descending)
    qsort(charFreqs, count, sizeof(CharFreq), compare);
    
    // Calculate total presses optimally
    int totalPresses = 0;
    for (int i = 0; i < count; i++) {
        int position = i / 8 + 1;
        totalPresses += charFreqs[i].freq * position;
    }
    
    return totalPresses;
}

int main() {
    char word[100001];
    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 the word to count characters

✓ Linear Growth

Space Complexity

O(1)

Fixed space for character frequency map (at most 26 characters)

✓ Linear Space

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Minimum Number of Pushes to Type Word I - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Character Counting — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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