Amazon Interview Question for Software Engineer / Developers


Country: United States
Interview Type: Phone Interview




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0
of 0 vote

what if everything is zero?

- Anonymous April 11, 2017 | Flag Reply
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0
of 0 vote

Not necessarily... The question says, k swap operations are allowed. If you closely look at the solution, the answer to the input is 9 and not 10. But your code will return 10 as the longest subarray starting with start_pos=1 and end_pos=10. Consider another case, [0,0,0,0,0], k=3. Your code would return 3. But the actual answer is 0 as there are no 1s to swap with 0s.

- ssk April 11, 2017 | Flag Reply
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0
of 0 vote

Solution using Dynamic Programming

public static int maxSubsequenceAfterSwap(int[] arr, int k) {
		int count = 0;
		int startIndex = 0;
		int maxLen = 0, maxIndex = 0;
		Queue<Integer> positionQ = new LinkedList<Integer>();
		for (int i = 0; i < arr.length; i++) {
			if (arr[i] == 0) {
				if (count < k) {
					positionQ.add(i);
					count++;
				} else {
					if (i - startIndex > maxLen) {
						maxLen = i - startIndex;
						maxIndex = startIndex;
					}
					startIndex = positionQ.remove() + 1;
					positionQ.add(i);
				}
			}
		}

		print("Max len=" + maxLen + ", starting index=" + maxIndex);
		return maxLen;
	}

- Ag April 13, 2017 | Flag Reply
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0
of 0 vote

Solution using DP

public static int maxSubsequenceAfterSwap(int[] arr, int k) {
		int count = 0;
		int startIndex = 0;
		int maxLen = 0, maxIndex = 0;
		Queue<Integer> positionQ = new LinkedList<Integer>();
		for (int i = 0; i < arr.length; i++) {
			if (arr[i] == 0) {
				if (count < k) {
					positionQ.add(i);
					count++;
				} else {
					if (i - startIndex > maxLen) {
						maxLen = i - startIndex;
						maxIndex = startIndex;
					}
					startIndex = positionQ.remove() + 1;
					positionQ.add(i);
				}
			}
		}

		print("Max len=" + maxLen + ", starting index=" + maxIndex);
		return maxLen;

}

- Ag April 13, 2017 | Flag Reply
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0
of 0 vote

i didn't understood the question

- bapan April 14, 2017 | Flag Reply
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0
of 0 vote

Recursive brute force.

static int maxOneSequence(int[] array, int swapsAllowed)
        {
            if (swapsAllowed == 0)
            {
                return countSequenceLen(array);
            }

            int maxLen = 0;
            for (int i = 0; i < array.Length; i++)
            {
                if (array[i] == 1)
                {
                    for (int j = 0; j < array.Length; j++)
                    {
                        if (array[j] == 0)
                        {
                            int len =  maxOneSequence(createSwappedArray(array, i, j), swapsAllowed - 1);
                            if (len > maxLen) { maxLen = len; }
                        }
                    }
                }
            }
            return maxLen;

}

- Mika April 15, 2017 | Flag Reply
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-1
of 1 vote

public int maxSubarray(int[] arr,int k){
	int max = 0;
	int diff = 0;//num zeros
	int totalOnes = 0;
	for(int i = 0; i < arr.length; i++){
		if(arr[i] == 1){
			totalOnes++;
		}
	}
	int j = 0;
	int i = 0;
	while(j < arr.length){
		if(arr[j]  == 0){
			diff++;
		}else{
		totalOnes--;
			diff--;
		}
		while(diff > k){
			if(arr[i] == 1){
				diff++;
				totalOnes++;
			}else{
				diff--;
			}
		}
		if(totalOnes >= k){
			max = Math.max(max,j - i + 1);
			
		}
		j++;
		
	}
	return max;

}

- divm01986 April 12, 2017 | Flag Reply
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-2
of 2 vote

The answer is the length of the largest sliding window possible, with 'K' zeros.

You also need to handle the case, where number of 1s outside the window should be greater or equal to 'K'.

public static int getMaxLen(int[] arr, int k){
		int wL = 0, wR = 0;
		int n = 0;
		int maxLen = 0;
		int maxWL = 0, maxWR = 0;
		if(arr[wR] == 0) n =1;
		while(wR < arr.length-1){
			if(n <= k){
				wR++;
				if(arr[wR] == 0) n++;
			}
			if(n > k){
				if(arr[wL] == 0) n--;
				wL++;
			}
			
			if((wR - wL) > maxLen){
				maxLen = wR - wL;
				maxWL = wL;
				maxWR = wR;
			}
		}
		int oneC = 0;
		for(int i = 0; i<maxWL; i++){
			if(arr[i] == 1) oneC++;
		}
		for(int i = maxWR+1; i<arr.length; i++){
			if(arr[i] == 1) oneC++;
		}
		if(oneC >= k)
			return maxLen;
		else{
			return (maxLen - oneC);
		}
	}
	
	public static void main(String[] args){
		int[] arr= {0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0};
		System.out.println(getMaxLen(arr, 3));
	}

- Anon April 11, 2017 | Flag Reply


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