167. Two Sum II — Input Array Is Sorted (Leetcode || Java || Easy)

Given a 1-indexed array of integers numbers that is already sorted in non-decreasing order, find two numbers such that they add up to a specific target number. Let these two numbers be numbers[index1] and numbers[index2] where 1 <= index1 < index2 <= numbers.length.

Return the indices of the two numbers, index1 and index2, added by one as an integer array [index1, index2] of length 2.

The tests are generated such that there is exactly one solution. You may not use the same element twice.

Your solution must use only constant extra space.

Example 1:

Input: numbers = [2,7,11,15], target = 9
Output: [1,2]
Explanation: The sum of 2 and 7 is 9. Therefore, index1 = 1, index2 = 2. We return [1, 2].

Example 2:

Input: numbers = [2,3,4], target = 6
Output: [1,3]
Explanation: The sum of 2 and 4 is 6. Therefore index1 = 1, index2 = 3. We return [1, 3].

Example 3:

Input: numbers = [-1,0], target = -1
Output: [1,2]
Explanation: The sum of -1 and 0 is -1. Therefore index1 = 1, index2 = 2. We return [1, 2].

Constraints:

• 2 <= numbers.length <= 3 * 104
• -1000 <= numbers[i] <= 1000
• numbers is sorted in non-decreasing order.
• -1000 <= target <= 1000
• The tests are generated such that there is exactly one solution.

SOLUTION:

class Solution {
public int[] twoSum(int[] numbers, int target) {
for(int i = 0; i <= numbers.length; i++){//We are going to run the loop for every single number
int num1 = numbers[i];//Consider number at index i as the first number we want

//Range for searching the second dnuber
int start = i + 1;
int end = numbers.length — 1;

while(start <= end){//Apply binary search for second number
int mid = start + (end — start)/2;
int ans = num1 + numbers[mid];

if(ans == target){//If we get our target return he index + 1 of both numbers as in the question we are given a 1-indexed array
return new int[] {i + 1,mid + 1};
}
else if(ans > target){
end = mid — 1;
}else{
start = mid + 1;
}
}
}
return new int[] {-1,-1};//If no index found for target return {-1,-1}
}
}

TIME COMPLEXITY :- O(N)

SPACE COMLEXITY :- O(1)

Thank you for reading. If you have queries please let me know in the comment section.