167. Two Sum II - Input Array Is Sorted
Explanation
To solve this problem, we can use a two-pointer approach. Since the input array is sorted, we can start with two pointers - one at the beginning of the array and one at the end. We check the sum of the elements pointed to by these pointers. If the sum is equal to the target, we return the indices. If the sum is less than the target, we move the left pointer to the right to increase the sum. If the sum is greater than the target, we move the right pointer to the left to decrease the sum. We continue this process until we find the target sum.
- Time complexity: O(n) where n is the number of elements in the array.
- Space complexity: O(1) since we are using constant extra space.
class Solution {
public int[] twoSum(int[] numbers, int target) {
int left = 0;
int right = numbers.length - 1;
while (left < right) {
int sum = numbers[left] + numbers[right];
if (sum == target) {
return new int[]{left + 1, right + 1};
} else if (sum < target) {
left++;
} else {
right--;
}
}
return new int[]{-1, -1}; // No solution found
}
}Code Editor (Testing phase)
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