easy binary search two pointers

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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.

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Author: Gauranshgoel123

two-sum-ii-input-array-is-sorted java.java

@@ -0,0 +1,18 @@
+class Solution {
+    public int[] twoSum(int[] nums, int target) {
+        int l=0;
+        int r=nums.length-1;
+        while(l<r){
+            if(nums[l]+nums[r]==target){
+                return new int[]{l+1,r+1};
+            }
+            else if(nums[l]+nums[r]>target){
+                r--;
+            }
+            else{
+                l++;
+            }
+        }
+        return new int[]{-1,-1};
+    }
+}