Plus One LeetCode Solution | Easy Approach

Minimum Cost to Merge Stones
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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.

 Plus One Solutions

Time: O(n)
Space: O(1)

C++

class Solution {
 public:
  vector<int> plusOne(vector<int>& digits) {
    for (int i = digits.size() - 1; i >= 0; --i) {
      if (digits[i] < 9) {
        ++digits[i];
        return digits;
      }
      digits[i] = 0;
    }

    digits.insert(begin(digits), 1);
    return digits;
  }
};

Java

 class Solution {
  public int[] plusOne(int[] digits) {
    for (int i = digits.length - 1; i >= 0; i--) {
      if (digits[i] < 9) {
        ++digits[i];
        return digits;
      }
      digits[i] = 0;
    }

    int[] ans = new int[digits.length + 1];
    ans[0] = 1;
    return ans;
  }
}

Python

class Solution:
  def plusOne(self, digits: List[int]) -> List[int]:
    for i, d in reversed(list(enumerate(digits))):
      if d < 9:
        digits[i] += 1
        return digits
      digits[i] = 0

    return [1] + digits

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