Integers: Plus one

Problem statements:

• leetcode-66

• You are given a large integer represented as an integer array digits, where each digits[i] is the ith digit of the integer. The digits are ordered from most significant to least significant in left-to-right order. The large integer does not contain any leading 0's.

• Increment the large integer by one and return the resulting array of digits.

• Example 1:

  • Input: digits = [1,2,3]
  • Output: [1,2,4]
  • Explanation:
    • The array represents the integer 123.
    • Incrementing by one gives 123 + 1 = 124.
    • Thus, the result should be [1,2,4].

• Example 2:

  • Input: digits = [4,3,2,1]
  • Output: [4,3,2,2]
  • Explanation:
    • The array represents the integer 4321.
    • Incrementing by one gives 4321 + 1 = 4322.
    • Thus, the result should be [4,3,2,2].

• Example 3:

  • Input: digits = [9]
  • Output: [1,0]
  • Explanation:
    • The array represents the integer 9.
    • Incrementing by one gives 9 + 1 = 10.
    • Thus, the result should be [1,0].

• Constraints:

  • 1 <= digits.length <= 100
  • 0 <= digits[i] <= 9
  • digits does not contain any leading 0's.

Solution:

• Appraoch:
  • We will need a array if all elements are 9 in the input array.
  • And in this case all other values in array will become 1.
• Implementation:

      public int[] plusOne(int[] digits) {
  
          for(int i = digits.length - 1; i >=0; i--) {
              if (digits[i] < 9) {
                  digits[i] += 1;
                  return digits;
              }
              digits[i] = 0;
          }
  
          // that is all digits are 9, new array with size + 1 and set MSB to '1'
          int[] newDigits = new int[digits.length + 1];
          newDigits[0] = 1;
  
          return newDigits;
      }