34. Search for a Range

Total Accepted: 91870
Total Submissions: 308813
Difficulty: Medium

Given a sorted array of integers, find the starting and ending position of a given target value.
Your algorithm's runtime complexity must be in the order of O(log n).
If the target is not found in the array, return [-1, -1]
.
For example,Given [5, 7, 7, 8, 8, 10]
and target value 8,return [3, 4]
.

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   // https://discuss.leetcode.com/topic/6327/a-very-simple-java-solution-with-only-one-binary-search-algorithm
    public int[] searchRange(int[] A, int target) {
        int start = firstGreaterEqual(A, target);
        if (start == A.length || A[start] != target) {
            return new int[]{-1, -1};
        }
        int second = firstGreaterEqual(A, target + 1) - 1;
        return new int[]{start, second};
    }

    //find the first number that is greater than or equal to target.
    //could return A.length if target is greater than A[A.length-1].
    //actually this is the same as lower_bound in C++ STL.
    private int firstGreaterEqual(int[] A, int target) {
        int low = 0, high = A.length;
        while (low < high) {
            int mid = low + ((high - low) >> 1);
            //low <= mid < high
            if (A[mid] < target) {
                low = mid + 1;
            } else {
                //should not be mid-1 when A[mid]==target.
                //could be mid even if A[mid]>target because mid<high.
                high = mid;
            }
        }
        return low;
    }