Add One to List - Amazon Top Interview Questions
Problem Statement :
You are given a list of integers nums, representing a decimal number and nums[i] is between [0, 9]. For example, [1, 3, 9] represents the number 139. Return the same list in the same representation except modified so that 1 is added to the number. Constraints 1 ≤ n ≤ 100,000 where n is the length of nums. Example 1 Input nums = [1, 9, 1] Output [1, 9, 2] Example 2 Input nums = [9] Output [1, 0]
Solution :
Solution in C++ :
vector<int> solve(vector<int>& nums) {
if (nums.empty()) return nums;
int i = nums.size() - 1;
while (nums[i] == 9 && i != 0) {
nums[i] = 0;
i--;
}
if (nums.size() == 100000 && nums[i] == 9) return nums;
if (nums[i] == 9) {
nums[i] = 1;
nums.push_back(0);
} else
nums[i] += 1;
return nums;
}
Solution in Java :
import java.util.*;
class Solution {
public int[] solve(int[] nums) {
if (nums[nums.length - 1] < 9) {
nums[nums.length - 1] += 1;
return nums;
}
int[] rtn = new int[nums.length + 1];
int carry = 1;
for (int i = nums.length - 1; i >= 0; i--) {
if (nums[i] < 9) {
nums[i] += carry;
return nums;
} else
nums[i] = 0;
}
rtn[0] = 1;
return rtn;
}
}
Solution in Python :
class Solution:
def solve(self, nums):
i = len(nums) - 1
while i >= 0:
if nums[i] + 1 <= 9:
nums[i] = nums[i] + 1
break
else:
nums[i] = 0
i -= 1
if i < 0:
nums.insert(0, 1)
return nums
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