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If the 3D array is sorted in all the 3 directions, its 1D equivalent array would also be sorted. If A[L][M][N] is a sorted 3D array, then A1[L*M*N], where A[i][j][k] = A1[i*M*N+j*N+k] will be a sorted 1 D array of (L*M*N) elements. The complexity of binary search on this would be log(L*M*N).
- ynsairam August 21, 2012Please correct me if wrong.