Opposite ends (Two Pointers Sub-pattern)

// Opposite ends on a sorted array: find a pair with the given sum.
// Each step throws away one element that can never be part of the answer.
pair<int, int> pairWithSum(vector<int>& a, int target) {
    int left = 0, right = (int)a.size() - 1;
    while (left < right) {
        int sum = a[left] + a[right];
        if (sum == target) return {left, right};
        if (sum < target) left++;   // need bigger: a[left] is too small for every partner
        else right--;               // need smaller: a[right] is too big for every partner
    }
    return {-1, -1};
}