Coder's Cat

LeetCode: Two Sum II - Input array is sorted

2020-03-16

Challenge Description: Two Sum II - Input array is sorted

Note the input array is already sorted. So we can iterate all the elements, for each element(suppose with a value of v1), we try to find another element with the value of target-v1. Binary search will suitable for this task.

The overall time complexity is $O(NlogN)$.

class Solution {
public:
vector<int> twoSum(vector<int>& numbers, int target) {
for(int i=0 ;i < numbers.size() - 1; i++) {
int l = i + 1;
int r = numbers.size() - 1;
int t = target - numbers[i];
while(l <= r) {
int m = (l + r) / 2;
if(numbers[m] == t) {
return vector<int> { i+1, m+1 };
} else if (numbers[m] > t) {
r = m - 1;
} else {
l = m + 1;
}
}
}
return vector<int>{};
}
};

Approach #2: two-slice

The idea of two slices is similar to binary search. In this approach, we use two index, index l is from left to right, index r is from right to left.

In each iteration, we compare the current sum of two elements with the target. There are tree cases:

  • sum == target, then we need return current two indexes.
  • sum > target, then we need to decrease index r .
  • sum < target, then we need to increase index l.

    The time complexity is $O(N)$.

    class Solution {
    public:
    vector<int> twoSum(vector<int>& numbers, int target) {
    int l = 0;
    int r = numbers.size()-1;
    while(l < r) {
    int s = numbers[l] + numbers[r];
    if(s == target) return vector<int> {l+1, r+1};
    if(s > target) r--;
    if(s < target) l++;
    }
    return vector<int>{};
    }
    };