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Leetcode 120 - Triangle, the first discussion
Study Leetcode Triangle, go over the note first, and also organize the notes to fit my reading style.
*** reasoning using dynamic programming ***
The triangle has a tree-like structure, which would be related to traversal algorithms such as DFS.
However the adjacent nodes always share a ‘branch’. In other word, there are overlapping subproblems.
Also, suppose x and y are ‘children’ of k. Once minimum paths from x and y to the bottom are known,
the minimum path starting from k can be decided in O(1), that is optimal substructure. Therefore,
dynamic programming would be the best solution to this problem in terms of time complexity.
*** top-down vs bottom-up debate ****
Let us look into the difference between ‘top-down’ and ‘bottom-up’ DP can be ‘literally’ pictured in the
input triangle.
For ‘top-down’ DP, starting from the node on the very top, we recursively find the minimum
path sum of each node. When a path sum is calculated, we store it in an array (memoization); the next time
we need to calculate the path sum of the same node, just retrieve it from the array. However, a cache that
is at least the same size as the input triangle itself is needed to store the pathsum, which takes O(N^2)
space. The space can be optimized, but the order of the nodes being processed is not straightforwardly seen
in a recursive solution, so deciding which part of the cache to discard can be a hard job.
‘Bottom-up’ DP, on the other hand, is very straightforward: we start from the nodes on the bottom row; the
min pathsums for these nodes are the values of the nodes themselves. From there, the min pathsum at the ith
node on the kth row would be the lesser of the pathsums of its two children plus the value of itself, i.e.:
minpath[k][i] = min( minpath[k+1][i], minpath[k+1][i+1]) + triangle[k][i];
Or even better, since the row minpath[k + 1] would be useless after minpath[k] is computed, we can simply set
minpath as a 1D array, and iteratively update itself:
For the kth level:
minpath[i] = min( minpath[i], minpath[i+1]) + triangle[k][i];
Thus, we have the following solution
int minimumTotal(vector<vector<int> > &triangle) {
int n = triangle.size();
vector<int> minlen(triangle.back());
for (int layer = n-2; layer >= 0; layer--) // For each layer
for (int i = 0; i <= layer; i++) // Check its every 'node'
// Find the lesser of its two children, and sum the current value in the triangle with it.
minlen[i] = min(minlen[i], minlen[i+1]) + triangle[layer][i];
return minlen[0];
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