Two Sum II: the idea to remember.

The core insight

Compare the endpoints: move left for a sum that is too small and right for a sum that is too large.

Why it works

Sorted order lets us discard an endpoint: if even its largest partner is too small, that left value cannot work.

The mistake to avoid

The output is one based. The pointer movement relies on sorted input.

Commented Java solution

Use java.util.* imports and a surrounding solution class. Tree snippets assume the standard TreeNode definition; linked lists use ListNode and Clone Graph uses a Node with val and neighbors; Copy Random List uses Node with val, next and random. Read the input assumptions above.

int[] twoSum(int[] nums, int target) {
    int left = 0, right = nums.length - 1;
    while (left < right) {
        long sum = (long) nums[left] + nums[right];
        if (sum == target) return new int[]{left + 1, right + 1};
        if (sum < target) left++;
        else right--;
    }
    throw new IllegalArgumentException("No valid pair");
}

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