https://codility.com/programmers/lessons/90-tasks_from_indeed_prime_2015_challenge/flood_depth/
https://codesays.com/2016/solution-to-flood-depth-by-codility/
This is a variant of Trapping Rain Water by LeetCode. The undergoing concept is the same. However, the implementation is a bit different.
http://stackoverflow.com/questions/12329711/interview-test-deepest-pit
https://lehoangquyenblog.wordpress.com/codility-lessons-solutions/90-flooddepth/
https://github.com/RIP21/Playground/blob/master/java/src/main/java/com/los/codility/FloodDepth.java
You are helping a geologist friend investigate an area with mountain lakes. A recent heavy rainfall has flooded these lakes and their water levels have reached the highest possible point. Your friend is interested to know the maximum depth in the deepest part of these lakes.
We simplify the problem in 2-D dimensions. The whole landscape can be divided into small blocks and described by an array A of length N. Each element of A is the altitude of the rock floor of a block (i.e. the height of this block when there is no water at all). After the rainfall, all the low-lying areas (i.e. blocks that have higher blocks on both sides) are holding as much water as possible. You would like to know the maximum depth of water after this entire area is flooded. You can assume that the altitude outside this area is zero and the outside area can accommodate infinite amount of water.
For example, consider array A such that:
A[0] = 1
A[1] = 3
A[2] = 2
A[3] = 1
A[4] = 2
A[5] = 1
A[6] = 5
A[7] = 3
A[8] = 3
A[9] = 4
A[10] = 2
The following picture illustrates the landscape after it has flooded:
The gray area is the rock floor described by the array A above and the blue area with dashed lines represents the water filling the low-lying areas with maximum possible volume. Thus, blocks 3 and 5 have a water depth of 2 while blocks 2, 4, 7 and 8 have a water depth of 1. Therefore, the maximum water depth of this area is 2.
Write a function:
int solution(int A[], int N);
that, given a non-empty zero-indexed array A consisting of N integers, returns the maximum depth of water.
Given array A shown above, the function should return 2, as explained above.
For the following array:
A[0] = 5
A[1] = 8
the function should return 0, because this landscape cannot hold any water.
Assume that:
- N is an integer within the range [1..100,000];
- each element of array A is an integer within the range [1..100,000,000].
Complexity:
- expected worst-case time complexity is O(N);
- expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).
Elements of input arrays can be modified.
https://codesays.com/2016/solution-to-flood-depth-by-codility/
This is a variant of Trapping Rain Water by LeetCode. The undergoing concept is the same. However, the implementation is a bit different.
def solution(A):
if len(A) < 3:
return 0
# The blocks in the stack is in strictly height descending order.
# For the first block in the stack, its max_depth is maximum water
# depth of its (exclusive) left area.
# The other blocks' max_depth is the maximum water depth between its
# previous block in the stack and itself, both exclusive.
Block = namedtuple("Block", ["height", "max_depth"])
stack = [Block(A[0],0)]
for height in A[1:]:
if height == stack[-1].height:
# These two adjacent blocks have the same height. They act
# totally the same in building any water container.
continue
elif height < stack[-1].height:
stack.append(Block(height, 0))
else:
max_depth = 0
# Let the current iterating block be C, the previous two
# blocks in the stack be A and B. And their positions are
# demoed as:
# C
# A C
# A ... B ... C
# while the blocks between A and B are omitted. So do the
# blocks between B and C.
#
# The additional_depth consider the blocks A, B, and C only,
# and igonres all the omitted blocks, such as:
# C
# A C
# A B C (no block is between A and B, or B and C)
#
# HOWEVER, the additional_depth is not always the maximum
# water depth between A and C, because there may be some
# water between A and B, or B and C, as exists in the omitted
# blocks. We need to adjust the additional_depth to get the
# maximum water depth between A and C, both exclusive.
while len(stack) > 1 and height > stack[-1].height:
additional_depth = min(stack[-2].height, height) - \
stack[-1].height
max_depth = max(max_depth, stack[-1].max_depth) + \
additional_depth
stack.pop()
# Combine leftward same-or-less-height blocks. These dropped
# blocks are never going to be part of the remaining water
# container.
while len(stack) > 0 and height >= stack[-1].height:
max_depth = max(max_depth, stack[-1].max_depth)
stack.pop()
stack.append(Block(height, max_depth))
overall_max_depth = 0
for block in stack:
if block.max_depth > overall_max_depth:
overall_max_depth = block.max_depth
return overall_max_depth
int solution(vector<int> &A) {
int left = 0, right = A.size() - 1, maxHeightFromLeft = 0, maxHeightFromRight = 0, ans = 0;
while(left < right) {
if(A[left] <= A[right]) {
maxHeightFromLeft = std::max(maxHeightFromLeft, A[left]);
ans = std::max(ans, maxHeightFromLeft - A[left++]);
}
else {
maxHeightFromRight = std::max(maxHeightFromRight, A[right]);
ans = std::max(ans, maxHeightFromRight - A[right--]);
}
}
return ans;
}
http://stackoverflow.com/questions/12329711/interview-test-deepest-pit
https://lehoangquyenblog.wordpress.com/codility-lessons-solutions/90-flooddepth/
int solution(vector<int> &A) { // write your code in C++14 (g++ 6.2.0) vector<int> highest_right_wall(A.size()); int higest_right = 0; if (A.size() > 1) { for (ssize_t i = A.size() - 2; i >= 0; --i) { highest_right_wall[i] = higest_right = max(higest_right, A[i + 1]); } } #if DEBUG LOG("highest_right_wall: "); for (auto h : highest_right_wall) { LOG("%d ", h); } LOG("\n"); #endif int highest_depth = 0; int current_water_top = 0; for (size_t i = 0; i < A.size(); ++i) { LOG("A[%d]=%d, right wall=%d\n", (int)i, A[i], highest_right_wall[i]); if (A[i] > current_water_top) { current_water_top = min(A[i], highest_right_wall[i]);//start of new water area } else { current_water_top = min(current_water_top, highest_right_wall[i]);//start of new water area int depth = 0; if (current_water_top > A[i]) depth = current_water_top - A[i]; highest_depth = max(highest_depth, depth); LOG("--> depth = %d, highest = %d\n", depth, highest_depth); } LOG("current_water_top=%d\n", current_water_top); } return highest_depth;}https://github.com/RIP21/Playground/blob/master/java/src/main/java/com/los/codility/FloodDepth.java
public int getMaximumDepth(int[] A) {
int highestIdx = 0;
int lowestIdx = 0;
int max = 0;
for (int i = 1; i < A.length; i++) {
int current = A[i];
int highest = A[highestIdx];
int lowest = A[lowestIdx];
if (current > highest) {
max = Math.max(highest - lowest, max);
highestIdx = i;
lowestIdx = highestIdx;
} else if (current > lowest) {
max = Math.max(current - lowest, max);
} else if (current < lowest) {
lowestIdx = i;
}
}
return max;
}
