Amazon Interview Question for Software Developers


Country: India




Comment hidden because of low score. Click to expand.
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of 0 vote

int maxWater(int[] arr)
{
	if(arr == null || arr.length < 3) {
		return -1;
	}

	int leftKeep = 0;
	int maxWater = 0;
	int leftCol = 0;
	for (int i = 1; i < arr.length - 1; i++) {
		// left + curr removed
		
		int maxRmvCurr = Math.max(Math.min(leftCol, arr[i + 1]) * 2, leftKeep + Math.min(arr[i - 1], arr[i + 1]));
		maxWater = Math.max (maxRmvCurr, maxWater);
		leftKeep = Math.max(leftKeep,Math.min(arr[i],leftCol));
		leftCol = arr[ i - 1];
	}

	return maxWater;

}

- divm01986 October 20, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
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of 0 vote

1. Compute leftmax and rightmax arrays. leftmax[i] denotes max in arr[0..i]. rightmax[i] denotes max in arr[i..n-1].
2. compute extra[] for each tower.
extra[i] = ( (min(leftmax[i], rightmax[i]) - arr[i] > 0) ? arr[i] : min(leftmax[i], rightmax[i]))
3. compute the ans as in the normal algorithm.
ans += ( (min(leftmax[i], rightmax[i]) - arr[i] > 0) ? min(leftmax[i], rightmax[i]) - arr[i] > 0 : 0)
4. now the final result would be ans + firstmax(extra) + secondmax(extra).
Please let me know if there is anything wrong in this approach.

- Raghu October 21, 2018 | Flag Reply


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