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- Find an element i > 0, so that s[i] < s[i-1] (i.e. the first pair of adjacent elements of s where the second element is smaller than the previous one).
- Lillienware February 07, 2015- Then sort the rest of the array ( s[i] to s[s.size() - 1] ) in descendant order.
- If you don't find that pair in the whole array, it means that s is already sorted in ascendant order. If "pointy" means that the array has to have a "descending part" at the end (k has to be less than s.size() - 1), you can just swap the last two elements of s (then k will be s.size()-2).