714253. Special Digits

Hard

Special Digits

Slug: special-digits--170647 Difficulty: Hard Id: 714253 Topic Tags: Combinatorial, Mathematical Company Tags: None

Summary

The Special Digits problem is a combinatorial challenge that requires finding the number of digits in a given integer that satisfy certain conditions. The key concept involved is combinatorics and mathematical manipulations.

Detailed Explanation

Given an integer n, we need to find the count of special digits in it. A digit is considered special if it appears at least once in each of the following three positions: hundreds, tens, and ones places. For example, for the number 123, the special digits are 1 and 3 because they appear in each of the three positions.

To solve this problem, we can use a simple algorithmic approach:

  1. Convert the integer n to a string to easily manipulate its digits.
  2. Initialize three counters (hundreds, tens, and ones) to keep track of the special digits found in each position.
  3. Iterate through the digits of the integer, considering each digit for the hundreds place (rightmost), tens place (second rightmost), and ones place (leftmost).
  4. For each digit, check if it appears in all three positions. If yes, increment the corresponding counter (hundreds, tens, or ones).
  5. After iterating through all digits, return the minimum of the three counters as the count of special digits.

Here's a step-by-step breakdown of the solution:

  n = 123
  str_n = "123"
  
  hundreds = 0
  tens = 0
  ones = 0
  
  for i in range(len(str_n)):
    digit = int(str_n[i])
    
    if (i == len(str_n) - 1):      // hundreds place
      hundreds += 1
    elif (i == len(str_n) - 2):   // tens place
      tens += 1
    else:                        // ones place
      ones += 1
  
  return min(hundreds, tens, ones)

Time complexity: O(log n), where n is the input integer. This is because we need to iterate through each digit of the integer, and the number of digits grows logarithmically with the value of the integer.

Space complexity: O(1), as we only use a constant amount of space to store the counters and other variables.

Optimized Solutions

Java

View on GeeksforGeeks
public int specialDigits(int n) {
    String str_n = Integer.toString(n);
    
    int hundreds = 0;
    int tens = 0;
    int ones = 0;
    
    for (int i = str_n.length() - 1; i >= 0; i--) {
        char digit = str_n.charAt(i);
        
        if (i == str_n.length() - 1) {      // hundreds place
            hundreds += 1;
        } else if (i == str_n.length() - 2) {   // tens place
            tens += 1;
        } else {                        // ones place
            ones += 1;
        }
    }
    
    return Math.min(hundreds, Math.min(tens, ones));
}

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