Cody

# Problem 11. Back and Forth Rows

Solution 2055190

Submitted on 12 Dec 2019 by Sumit Gavasane
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### Test Suite

Test Status Code Input and Output
1   Pass
n = 4; a = [ 1 2 3 4; 8 7 6 5; 9 10 11 12; 16 15 14 13]; assert(isequal(a,back_and_forth(n)));

b = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 sa = 4 4 sar = 4 sac = 4 z = uint32 1 k = -1 b = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 z = uint32 1 k = -1 b = 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 z = uint32 1 k = -1 b = 1 2 3 1 1 1 1 1 1 1 1 1 1 1 1 1 z = uint32 1 k = -1 b = 1 2 3 4 1 1 1 1 1 1 1 1 1 1 1 1 z = uint32 1 k = 0 b = 1 2 3 4 8 1 1 1 1 1 1 1 1 1 1 1 z = uint32 1 k = 0 b = 1 2 3 4 8 7 1 1 1 1 1 1 1 1 1 1 z = uint32 1 k = 0 b = 1 2 3 4 8 7 6 1 1 1 1 1 1 1 1 1 z = uint32 1 k = 0 b = 1 2 3 4 8 7 6 5 1 1 1 1 1 1 1 1 z = uint32 2 k = -1 b = 1 2 3 4 8 7 6 5 9 1 1 1 1 1 1 1 z = uint32 2 k = -1 b = 1 2 3 4 8 7 6 5 9 10 1 1 1 1 1 1 z = uint32 2 k = -1 b = 1 2 3 4 8 7 6 5 9 10 11 1 1 1 1 1 z = uint32 2 k = -1 b = 1 2 3 4 8 7 6 5 9 10 11 12 1 1 1 1 z = uint32 2 k = 0 b = 1 2 3 4 8 7 6 5 9 10 11 12 16 1 1 1 z = uint32 2 k = 0 b = 1 2 3 4 8 7 6 5 9 10 11 12 16 15 1 1 z = uint32 2 k = 0 b = 1 2 3 4 8 7 6 5 9 10 11 12 16 15 14 1 z = uint32 2 k = 0 b = 1 2 3 4 8 7 6 5 9 10 11 12 16 15 14 13

2   Pass
n = 5; a = [ 1 2 3 4 5; 10 9 8 7 6; 11 12 13 14 15; 20 19 18 17 16; 21 22 23 24 25]; assert(isequal(a,back_and_forth(n)));

b = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 sa = 5 5 sar = 5 sac = 5 z = uint32 1 k = -1 b = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 z = uint32 1 k = -1 b = 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 z = uint32 1 k = -1 b = 1 2 3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 z = uint32 1 k = -1 b = 1 2 3 4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 z = uint32 1 k = -1 b = 1 2 3 4 5 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 z = uint32 1 k = 0 b = 1 2 3 4 5 10 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 z = uint32 1 k = 0 b = 1 2 3 4 5 10 9 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 z = uint32 1 k = 0 b = 1 2 3 4 5 10 9 8 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 z = uint32 1 k = 0 b = 1 2 3 4 5 10 9 8 7 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 z = uint32 1 k = 0 b = 1 2 3 4 5 10 9 8 7 6 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 z = uint32 2 k = -1 b = 1 2 3 4 5 10 9 8 7 6 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 z = uint32 2 k = -1 b = 1 2 3 4 5 10 9 8 7 6 11 12 1 1 1 1 1 1 1 1 1 1 1 1 1 z = uint32 2 k = -1 b = 1 2 3 4 5 10 9 8 7 6 11 12 13 1 1 1 1 1 1 1 1 1 1 1 1 z = uint32 2 k = -1 b = 1 2 3 4 5 10 9 8 7 6 11 12 13 14 1 1 1 1 1 1 1 1 1 1 1 z = uint32 2 k = -1 b = 1 2 3 4 5 10 9 8 7 6 11 12 13 14 15 1 1 1 1 1 1 1 1 1 1 z = uint32 2 k = 0 b = 1 2 3 4 5 10 9 8 7 6 11 12 13 14 15 20 1 1 1 1 1 1 1 1 1 z = uint32 2 k = 0 b = 1 2 3 4 5 10 9 8 7 6 11 12 13 14 15 20 19 1 1 1 1 1 1 1 1 z = uint32 2 k = 0 b = 1 2 3 4 5 10 9 8 7 6 11 12 13 14 15 20 19 18 1 1 1 1 1 1 1 z = uint32 2 k = 0 b = 1 2 3 4 5 10 9 8 7 6 11 12 13 14 15 20 19 18 17 1 1 1 1 1 1 z = uint32 2 k = 0 b = 1 2 3 4 5 10 9 8 7 6 11 12 13 14 15 20 19 18 17 16 1 1 1 1 1 z = uint32 3 k = -1 b = 1 2 3 4 5 10 9 8 7 6 11 12 13 14 15 20 19 18 17 16 21 1 1 1 1 z = uint32 3 k = -1 b = 1 2 3 4 5 10 9 8 7 6 11 12 13 14 15 20 19 18 17 16 21 22 1 1 1 z = uint32 3 k = -1 b = 1 2 3 4 5 10 9 8 7 6 11 12 13 14 15 20 19 18 17 16 21 22 23 1 1 z = uint32 3 k = -1 b = 1 2 3 4 5 10 9 8 7 6 11 12 13 14 15 20 19 18 17 16 21 22 23 24 1 z = uint32 3 k = -1 b = 1 2 3 4 5 10 9 8 7 6 11 12 13 14 15 20 19 18 17 16 21 22 23 24 25