## Interview Question

## Interview Answer

13 Answers

O(logn) approach: int max_value_in_semisorted_array(int* arr, int start, int end) { if (start == end) return arr[start]; int middle = (end + start) / 2; if (arr[middle] > arr[start]) return max_value_in_semisorted_array(arr, middle, end); else return max_value_in_semisorted_array(arr, start, middle); }

No, your O(logn)-approach doesn't work. Try this: 1 5 8 10 9 8 7 6 5 4

The previous O(log n) approach doesn't handle arrays of odd length. int findPeak(int * array, int start, int end) { if ((start == end) || ((start == (end - 1)) && (array[start] > array[end]))) return array[start]; if ((start == (end - 1)) && (array[start] > array[end])) return array[end]; int middle = floor((start + end) / 2); if (array[middle] > array[start]) return (findPeak(array, middle, end)); else return (findPeak(array, start, middle)); } Remember to include math.h for the floor function.

It seems, that it doesn't matter if the array is of odd or even length(btw, my example is not odd :)). This approach doesn't work when the max element is in the first half of your array and the middle element is greater than the first one. We cannot guarantee that if the middle element is greater than the first, our array increases monotonically from x[1] to x[middle] and we should search the max element in the range from middle to end.

It is necessary to check two elements on each side of our middle element to find out where the array increases , where decreases, and where the peak is. And only then we can guarantee something about our max.

The maximum is obviously just the inflection point in the list... some of the posted solutions here are absurdly complex for this.

Brute force solution can be ofcourse O(N). But we can make it better: mid = (start + end )/2 if mid > mid + 1 && mid > mid - 1 return mid else if mid < mid + 1 && mid > mid -1 search mid -> end else if mid > mid + 1 && mid < mid - 1 search start -> mid

@Mo - This won't be work for an array where the max element is hidden in the first / second half of the main array. Eg: {1,3,9,1,-2,6,2,2,5,-1}

In your test case, I see that the numbers increase, decrease, increases, decreases and so on. I thought the question says the list is semi sorted. So my program assumes that the numbers increase and then decrease. We need to find the peak. At least that is my understanding.

private static int recFindPeak(int[] a, int start, int end) { if(start == end) return a[start]; int mid = (start+end)/2; if((a[mid] > a[mid+1])&&(a[mid] > a[mid-1])) return a[mid]; else if(a[mid] > a[mid-1]) return recFindPeak(a, mid+1, end); else return recFindPeak(a, start, mid-1); }

My codes: int findPeak(vector<int> num) { if (num.empty()) return 0; int lower = 0; int upper = num.size()-1; while (lower <= upper) { if (upper-lower <= 1) return max(num[lower],num[upper]); int mid = lower+(upper-lower)/2; if (num[mid-1] < num[mid] && num[mid] > num[mid+1]) return num[mid]; else if (num[mid-1] < num[mid] && num[mid] < num[mid+1]) lower = mid+1; else upper = mid-1; } }

1. Walk from begin to mid comparing a[i] < a[mid] and stop incrementing i when this condition fails. 2. Walk from end to mid comparing a[mid] > a[j] and stop decrementing j when this condition fails. 3. If a[i] < a[j], then return a[j], else return a[i].

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Simply walk the array to find the max in O(n) time. Compare the current max to the next element all the way until the end of the array.