33 lines
1.1 KiB
C++
33 lines
1.1 KiB
C++
#include <vector>
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#include <algorithm>
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#include <numeric>
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/**
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* 883. Projection Area of 3D Shapes
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* You are given an n x n grid where we place some 1 x 1 x 1 cubes that are axis-aligned with the x, y, and z axes.
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* Each value v = grid[i][j] represents a tower of v cubes placed on top of the cell (i, j).
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* We view the projection of these cubes onto the xy, yz, and zx planes.
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* A projection is like a shadow, that maps our 3-dimensional figure to a 2-dimensional plane. We are viewing the "shadow" when looking at the cubes from the top, the front, and the side.
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* Return the total area of all three projections.
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*/
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class Solution {
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public:
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static int projectionArea(const std::vector<std::vector<int>>& G) {
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const int n = G.size();
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int xy = 0, yz = 0, zx = 0;
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std::for_each(G.begin(), G.end(), [&](const std::vector<int>& v) {
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xy += std::count_if(v.begin(), v.end(), [](int x) { return x > 0; });
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zx += *std::max_element(v.begin(), v.end());
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});
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for (int i = 0; i < n; ++i) {
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int t = 0;
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for (int j = 0; j < n; ++j) {
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t = std::max(t, G[j][i]);
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}
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yz += t;
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}
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return xy + yz + zx;
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}
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};
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