Find If Array Can Be Sorted

You are given a 0-indexed array of positive integers nums. In one operation, you can swap any two adjacent elements if they have the same number of set bits (1 presented in binary). You are allowed to do this operation any number of times (including zero). Return true if you can sort the array, else return false.

Examples:

Input: nums = [8, 4, 2, 30, 15] => [0x1000, 0x100, 0x10, 0x11110, 0x1111];
Output: True => Sorted to [0x10, 0x100, 0x1000, 0x1111, 0x11110] == [2, 4, 8, 15, 30];

Input: nums = [3, 16, 8, 4, 2];
Output: False;

Solution

class Solution {
public:
    static bool canSortArray(const vector<int>& nums) {
        uint16_t pmax = 0, cmin = 0; cmax = 0;
        uint8_t pcnt = 0;
        for (const uint16_t v : nums) {
            if (const uint8_t ccnt = popcount(v); pcnt == ccnt) {
                cmin = min(cmin, v);
                cmax = max(cmax, v);
            } else if (cmin < pmax) {
                return false;
            } else {
                pmax = cmax;
                cmin = cmax = v;
                pcnt = ccnt;
            }
        }
        return cmin >= pmax;
    }
}

Explanation:

This solution divides the array into several groups by the number of bits that are set. The popcount() function here is an library function that used to find the number of set bits in a variable. And the if (const uint8_t ccnt = popcount(v); pcnt == ccnt) structure is to define a variable within the if-else if-else structure and use it for boolean expression in a brief way. The if-else if-else structure in the code is doing mainly for:

• Checking set bits of current variable v, if current number of set bit ccnt equals the number of set bits of last variable pcnt, then they are in the same group. Then update the maximum and the minimum value of current group into cmin and cmax.
• If the minimum value of current group cmin is less than the greatest value of the last group pmax, then this array cannot be sorted because every number that presents in a relative back group must be greater than every number in a front group. Thus, program return false here.
• If ccnt does not equal pcnt, current value v belongs to a new group, so the maximum value of current group is assigned to pmax, the maximum of the previous group, then update cmin and cmax with current value, and pcnt with the number of set bit of v.

Above all, the core of this solution is to guarantee that for each group, their minimum value must be greater than the maximum value of every group in its front.