You can do it with 1D BIT too. Sort ladies by their increasing values of beauty. A lady is a self-murderer iff there is some lady who has a higher beauty, intellect and richness. Maintain a BIT which stores the maximum possible value of intellect for each value of richness. When you traverse the above sorted list in reverse order, if you're at position i, you know that the ladies inserted into the BIT already have higher beauty. Now find the maximum possible intellect for any value of richness greater than the richness of the current lady. If this found value > intellect value of current lady, then there is definitely some lady j (j > i in sorted list) who has higher values of all three parameters. Complexity: O(logn) per query, O(nlogn) overall. :)