Google Interview Question for Software Engineers


Country: United States
Interview Type: In-Person




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2
of 2 vote

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# The question is about identifying any window size k that its numbers at both ends bigger than numbers in between
def bloomDay(a, k): # k must be >= 0
    window = [] # elements in window maintain descending order. 
    firstDay = 10000000
    i = 0
    while i < len(a):
        if not window:
            window.append(i)
            i += 1
        elif i - window[0] <= k:
            if a[i] >= a[window[0]]:
                window = [i]
                i += 1
            elif a[i] >= a[window[-1]]:
                window.pop()
            else:
                window.append(i)
                i += 1
        else: #window size == k
            if len(window) == 1 or a[window[1]] < a[i]:#valid window with both end numbers biggest
                firstDay = min(firstDay, a[i], a[window[0]])
                window = [i]
                i += 1
            else:  #a[i] is less than a number in the middle of the window
                window.pop(0)
                if a[window[-1]] <= a[i]:
                    window.pop()
                else:
                    window.append(i)
                    i += 1
    return firstDay

- aonecoding January 05, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Can you explain the question with more details?

- rick January 08, 2018 | Flag Reply
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0
of 0 vote

int func(int[] arr, int k) {
        int ans = Integer.MAX_VALUE;
        Deque<Integer> q = new LinkedList<>();
        int ptr = 0;
        int length = arr.length;
        while (ptr < length) {
            int ele = arr[ptr];
            if (q.isEmpty() == true) {
                q.add(ptr);
                ptr++;
            } else {
                if (ptr-q.getFirst() <= k) {                    
                    if (ele >= arr[q.getFirst()]) {
                        while (q.isEmpty() == false) {
                            q.remove();
                        }
                        q.add(ptr);
                        ptr++;
                    } else {
                        if (ele >= arr[q.getLast()]) {
                            q.removeLast();
                        } else {
                            q.add(ptr);
                            ptr++;
                        }
                    }
                } else {
                    int x = arr[q.removeFirst()];
                    if (ele > arr[q.removeFirst()]) {
                        ans = Math.min(ans, Math.min(ele, x));
                        while (q.isEmpty() == false) {
                            q.remove();
                        }
                        q.add(ptr);
                        ptr++;
                    } else {
                        if (ele >= arr[q.getLast()]) {
                            q.removeLast();
                        } else {
                            q.add(ptr);
                            ptr++;
                        }
                    }
                }
            }
        }
        return ans;
    }

- 11.9.95.shubham January 09, 2018 | Flag Reply


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