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Leetcode 3014. Minimum Number of Pushes to Type Word I Sure! Let’s break down the problem and solve it systematically.

Problem Statement

You have a special keyboard with the following keys:

• Each alphabet letter ‘a’ to ‘e’.
• The backspace key to delete the last character.
• The confirm key to finish typing the current sequence.

You are initially located at ‘a’ on the keyboard. The cost to move to the next or previous letter in the sequence (circularly) is one push. You need to type a given word (string) using the minimum number of key pushes.

Function signature:

int minPushesToTypeWord(String word)

Clarifying Questions

1. What exactly does the input string contain?
   • The input consists only of lowercase letters from ‘a’ to ‘e’.
2. How do we handle backspace and confirm? Are these represented in the input?
   • The problem doesn’t specify inclusion of backspace or confirm directly in the string. We consider the input string as the target sequence we need to type.

Strategy

1. Understand Character Movement:
   • Movement is circular (‘a’ <-> ‘e’).
   • Moving from ‘a’ to ‘b’ or ‘e’ to ‘d’ is 1 step each. Moving from ‘a’ to ‘e’ is 1 step.
2. Compute the Minimum Steps:
   • For each character move, compute the minimum steps considering the circular nature.
3. Iterate through the String:
   • Track the current character (starting from ‘a’).
   • For each character in the string, calculate the minimum circular distance from the current to the target character.
   • Sum up these minimum distances.

Code Implementation

We’ll iterate through the given word and compute the total minimal pushes required.

public class MinimumPushes {

    public static int minPushesToTypeWord(String word) {
        int totalPushes = 0;
        char currentKey = 'a';
        
        for (int i = 0; i < word.length(); i++) {
            char targetKey = word.charAt(i);
            totalPushes += minStepsBetween(currentKey, targetKey);
            currentKey = targetKey;
        }
        
        return totalPushes;
    }

    private static int minStepsBetween(char from, char to) {
        int n = 5; // Given 'a', 'b', 'c', 'd', 'e'
        int fromPos = from - 'a'; // Position of 'from' in the sequence
        int toPos = to - 'a';     // Position of 'to' in the sequence
        return Math.min(Math.abs(toPos - fromPos), n - Math.abs(toPos - fromPos));
    }

    public static void main(String[] args) {
        String word = "cab";
        System.out.println(minPushesToTypeWord(word)); // Expected Output: 4
    }
}

Time Complexity

• Time Complexity: O(n), where n is the length of the string. We only iterate through the string once and do a constant amount of work (computing minimal steps) for each character.
• Space Complexity: O(1), as we are using only a few extra variables irrespective of the input size.

This solution should be efficient and straightforward for typing the given word using the fewest possible pushes on the special keyboard.

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