Hey, next problem from leetcode will be Triangle. Task description:
Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below. For example, given the following triangle
[ [2], [3,4], [6,5,7], [4,1,8,3] ]The minimum path sum from top to bottom is
11(i.e., 2 + 3 + 5 + 1 = 11). Note: Bonus point if you are able to do this using only O(n) extra space, where n is the total number of rows in the triangle.
Stop and think …
Judging to example we should go level by level and choose smallest “connected” value . Why I’ve written “connected” because if we have something like that
[5,6,7], [4,4,1,8]
we should choose 6,1 but not 5,4 because sum 6+1 < 5+4. OK interesting observation… So, what will be if we start from bottom to top ? We can choose smallest number and all will be correct. That is good, but how to check “connection”. In this situation we can use rule from Dynamic Programming(DP) (Store sum under i position of i-element from next level and min from i and i+1 element from stored values.) OK, lets go …
public int minimumTotal(@NotNull ListList<List<Integer>> triangle) {
if (triangle.size() == 0) return 0;
int[] cache = new int[triangle.size()];
List<Integer> bottom = triangle.get(triangle.size() - 1);
for (int i = 0; i < cache.length; i++) {
cache[i] = bottom.get(i);
}
int layer = cache.length - 2;
while (layer >= 0) {
List<Integer> list = triangle.get(layer);
for (int i = 0; i <= layer; i++) {
cache[i] = list.get(i) + Math.min(cache[i], cache[i + 1]);
}
layer--;
}
return cache[0];
}
What I’ve written :
Create cache (remember about DP)
int[] cache = new int[triangle.size()];
Get last line(level) from triangle and put to cache
List<Integer> bottom = triangle.get(triangle.size() - 1);
for (int i = 0; i < cache.length; i++) {
cache[i] = bottom.get(i);
}
after that start from next level and
Store sum under i position of i-element from next level and min from i and i+1 element from stored values.
while (layer >= 0) {
List<Integer> list = triangle.get(layer);
for (int i = 0; i <= layer; i++) {
cache[i] = list.get(i) + Math.min(cache[i], cache[i + 1]);
}
layer--;
}
How this technique is working ? That is very easy, try to test our example :
- firstly we have put last level to cache , so, cache contains next values
[4,1,8,3]
- start loop from next level
[6,5,7],
i == 0, so, we store under 0-position smallest sum (6+4 or 6+1)
cache[i] = list.get(i) + Math.min(cache[i], cache[i + 1]);
next positions the same, OK after second line we will have our cache like next
[7,6,8,3]
- go to next level
[3,4],
cache will be next
[9,9,8,3]
- go to next level
[2]
cache will be next
[11,9,8,3]
so, just return 0 position and that is all
return cache[0];
So, we solved next task, that is good 🙂 Thanks!