HCL Interview Question Interns
Team: 1
Country: India
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Not sure I understand the 4 cases, but here's my interpretation and solution.
- a triangle is a square block which contains a triangle in it, it can be of any size, (2x2, 3x3, 4x4...).
- a triangle is a block which has ones on one corner and zeros on the other corner... for example
is a triangle
- I excluded the triangles that have other stuff in the square like
Not a triangle. But not sure if it should be... from the question. Would need clarifications on this. If this is an acceptable triangle, we'd have the template be
[1,1,1
1,1,0
1,0,-1] where -1 is a don't care, shown later in the template making function. We really just need a solid triangle of 1s, and a line of 0 which is a boundary of the triangle, and the rest of the square, we don't really care.
This is the solution presented here as I find it's a bit more flexible and intuitive to someone looking for triangles.
My code uses somewhat of an image processing, template matching approach. I make 4 templates of triangles and convolve them on the matrix, counting everytime we find a match. I then increase the size of the triangle and convolve again, until the triangle is the size of the matrix, which I assumed to be square as stated in the question, but is not the case in the example.
public static int[,] MakeTriangleH(int n, int typeTriangle) { int[,] H = new int[n,n]; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) {// change the elseif to else if want strict triangles (remove else) if (typeTriangle == 0) {// top left triangle if ((n - i) > j) H[i, j] = 1; else if ((n - i) == j) H[i, j] = 0; else H[i, j] = -1; } if (typeTriangle == 1) {// top right triangle if (i <= j) H[i, j] = 1; else if (i == j+1) H[i, j] = 0; else H[i, j] = -1; } if (typeTriangle == 2) {// bottom right triangle if (i >= n - j - 1) H[i, j] = 1; else if (i == n - j-2) H[i, j] = 0; else H[i, j] = -1; } if (typeTriangle == 3) {// bottom left triangle H[i, j] = i >= j ? 1 : 0; if (i >= j) H[i, j] = 1; else if (i == j-1) H[i, j] = 0; else H[i, j] = -1; } } } return H; } public static bool PatternMatch(int [,] A, int iy, int ix, int [,] H) { for (int i = 0;i<H.GetLength(0);i++) { for (int j = 0;j<H.GetLength(0);j++) {// this check for -1 is in case the template contains a don't care if (H[i,j] != -1 && A[i+iy,j+ix] != H[i,j]) { return false; } } } return true; } public static int Convolve(int[,] A, int[,] H) { int count = 0; for (int i = 0; i <= A.GetLength(0) - H.GetLength(0); i++) { for (int j = 0; j <= A.GetLength(0) - H.GetLength(0); j++) { count += PatternMatch(A, i, j, H)?1:0; } } return count; } public static int CountTriangles(int[,] A) { int count = 0; int[,] H; for (int sizeTriangle = 2; sizeTriangle <= A.GetLength(0); sizeTriangle++) { H = MakeTriangleH(sizeTriangle, 0); count += Convolve(A, H); H = MakeTriangleH(sizeTriangle, 1); count += Convolve(A, H); H = MakeTriangleH(sizeTriangle, 2); count += Convolve(A, H); H = MakeTriangleH(sizeTriangle, 3); count += Convolve(A, H); } return count; }gives us 18 triangles
- djtrance January 31, 2016