Plus One Leetcode Solution
Plus One Leetcode Problem :
You are given a large integer represented as an integer array digits, where each digits[i] is the i th 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.
Plus One Leetcode Problem Solution :
Constraints :
- 1 <= digits.length <= 100
- 0 <= digits[i] <= 9
- digits does not contain any leading 0’s.
Example 1:
- 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 2:
- 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].
Idea:
Simple Brute Force Approach.
Approach:
Here, we are describing a process where we increment the last digit of an array outside of a loop, and then, within the loop, we have to check if any digit becomes 10.
If a digit becomes 10, you set it to 0 and increment the preceding digit by 1.
This process continues until there are no more 10s. Additionally, if the array has a single-digit number like 9, then add a 0 at the end and change it to 10.
Here’s the explanation for this process:
- Start with an array of digits.
- Increment the last digit by 1 outside of the loop.
- Enter a loop to handle carrying over when a digit becomes 10.
- Inside the loop, check if the current digit is 10.
- If it’s 10, set it to 0 and increment the preceding digit by 1.
- Continue this process until there are no more 10s.
- If the first digit becomes 10, add a new leading 1 to the array.
Finally, you have your modified array.
Complexity:
-
Time Complexity:
Time complexity is O(n). -
Space Complexity:
Space complexity is O(1).
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Code :
class Solution {
public:
vector plusOne(vector< int>& digits) {
int n=digits.size();
int sum=0;
digits[n-1]++;
for(int i=n-1;i>=0;i--){
if(digits[i]==10){
digits[i]=0;
if(i-1>=0)
{
digits[i-1]++;
}
else{
digits.push_back(0);
digits[i]++;
}
}
}
return digits;
}
};
import java.util.ArrayList;
import java.util.List;
class Solution {
public List< Integer> plusOne(List< Integer> digits) {
int n = digits.size();
int sum = 0;
digits.set(n-1, digits.get(n-1) + 1);
for (int i = n-1; i >= 0; i--) {
if (digits.get(i) == 10) {
digits.set(i, 0);
if (i-1 >= 0) {
digits.set(i-1, digits.get(i-1) + 1);
} else {
digits.add(0);
digits.set(i, digits.get(i) + 1);
}
}
}
return digits;
}
}class Solution:
def plusOne(self, digits):
n = len(digits)
digits[n-1] += 1
for i in range(n-1, -1, -1):
if digits[i] == 10:
digits[i] = 0
if i-1 >= 0:
digits[i-1] += 1
else:
digits.append(0)
digits[i] += 1
return digits
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