# 1387. Sort Integers by The Power Value

## Description

The power of an integer x is defined as the number of steps needed to transform x into 1 using the following steps:

• if x is even then x = x / 2
• if x is odd then x = 3 * x + 1

For example, the power of x = 3 is 7 because 3 needs 7 steps to become 1 (3 --> 10 --> 5 --> 16 --> 8 --> 4 --> 2 --> 1).

Given three integers lo, hi and k. The task is to sort all integers in the interval [lo, hi] by the power value in ascending order, if two or more integers have the same power value sort them by ascending order.

Return the kth integer in the range [lo, hi] sorted by the power value.

Notice that for any integer x (lo <= x <= hi) it is guaranteed that x will transform into 1 using these steps and that the power of x is will fit in a 32-bit signed integer.

Example 1:

Input: lo = 12, hi = 15, k = 2
Output: 13
Explanation: The power of 12 is 9 (12 --> 6 --> 3 --> 10 --> 5 --> 16 --> 8 --> 4 --> 2 --> 1)
The power of 13 is 9
The power of 14 is 17
The power of 15 is 17
The interval sorted by the power value [12,13,14,15]. For k = 2 answer is the second element which is 13.
Notice that 12 and 13 have the same power value and we sorted them in ascending order. Same for 14 and 15.


Example 2:

Input: lo = 7, hi = 11, k = 4
Output: 7
Explanation: The power array corresponding to the interval [7, 8, 9, 10, 11] is [16, 3, 19, 6, 14].
The interval sorted by power is [8, 10, 11, 7, 9].
The fourth number in the sorted array is 7.


Constraints:

• 1 <= lo <= hi <= 1000
• 1 <= k <= hi - lo + 1

## Solutions

Solution 1: Custom Sorting

First, we define a function $f(x)$, which represents the number of steps required to change the number $x$ to $1$, i.e., the weight of the number $x$.

Then, we sort all the numbers in the interval $[lo, hi]$ in ascending order of weight. If the weights are the same, we sort them in ascending order of the numbers themselves.

Finally, we return the $k$-th number after sorting.

The time complexity is $O(n \times \log n \times M)$, and the space complexity is $O(n)$. Where $n$ is the number of numbers in the interval $[lo, hi]$, and $M$ is the maximum value of $f(x)$. In this problem, the maximum value of $M$ is $178$.

• class Solution {
public int getKth(int lo, int hi, int k) {
Integer[] nums = new Integer[hi - lo + 1];
for (int i = lo; i <= hi; ++i) {
nums[i - lo] = i;
}
Arrays.sort(nums, (a, b) -> {
int fa = f(a), fb = f(b);
return fa == fb ? a - b : fa - fb;
});
return nums[k - 1];
}

private int f(int x) {
int ans = 0;
for (; x != 1; ++ans) {
if (x % 2 == 0) {
x /= 2;
} else {
x = x * 3 + 1;
}
}
return ans;
}
}

• class Solution {
public:
int getKth(int lo, int hi, int k) {
auto f = [](int x) {
int ans = 0;
for (; x != 1; ++ans) {
if (x % 2 == 0) {
x /= 2;
} else {
x = 3 * x + 1;
}
}
return ans;
};
vector<int> nums;
for (int i = lo; i <= hi; ++i) {
nums.push_back(i);
}
sort(nums.begin(), nums.end(), [&](int x, int y) {
int fx = f(x), fy = f(y);
if (fx != fy) {
return fx < fy;
} else {
return x < y;
}
});
return nums[k - 1];
}
};

• @cache
def f(x: int) -> int:
ans = 0
while x != 1:
if x % 2 == 0:
x //= 2
else:
x = 3 * x + 1
ans += 1
return ans

class Solution:
def getKth(self, lo: int, hi: int, k: int) -> int:
return sorted(range(lo, hi + 1), key=f)[k - 1]


• func getKth(lo int, hi int, k int) int {
f := func(x int) (ans int) {
for ; x != 1; ans++ {
if x%2 == 0 {
x /= 2
} else {
x = 3*x + 1
}
}
return
}
nums := make([]int, hi-lo+1)
for i := range nums {
nums[i] = lo + i
}
sort.Slice(nums, func(i, j int) bool {
fx, fy := f(nums[i]), f(nums[j])
if fx != fy {
return fx < fy
}
return nums[i] < nums[j]
})
return nums[k-1]
}

• function getKth(lo: number, hi: number, k: number): number {
const f = (x: number): number => {
let ans = 0;
for (; x !== 1; ++ans) {
if (x % 2 === 0) {
x >>= 1;
} else {
x = x * 3 + 1;
}
}
return ans;
};
const nums = new Array(hi - lo + 1).fill(0).map((_, i) => i + lo);
nums.sort((a, b) => {
const fa = f(a),
fb = f(b);
return fa === fb ? a - b : fa - fb;
});
return nums[k - 1];
}