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Formatted question description: https://leetcode.ca/all/2161.html

# 2161. Partition Array According to Given Pivot (Medium)

You are given a 0-indexed integer array nums and an integer pivot. Rearrange nums such that the following conditions are satisfied:

• Every element less than pivot appears before every element greater than pivot.
• Every element equal to pivot appears in between the elements less than and greater than pivot.
• The relative order of the elements less than pivot and the elements greater than pivot is maintained.
• More formally, consider every pi, pj where pi is the new position of the ith element and pj is the new position of the jth element. For elements less than pivot, if i < j and nums[i] < pivot and nums[j] < pivot, then pi < pj. Similarly for elements greater than pivot, if i < j and nums[i] > pivot and nums[j] > pivot, then pi < pj.

Return nums after the rearrangement.

Example 1:

Input: nums = [9,12,5,10,14,3,10], pivot = 10
Output: [9,5,3,10,10,12,14]
Explanation:
The elements 9, 5, and 3 are less than the pivot so they are on the left side of the array.
The elements 12 and 14 are greater than the pivot so they are on the right side of the array.
The relative ordering of the elements less than and greater than pivot is also maintained. [9, 5, 3] and [12, 14] are the respective orderings.


Example 2:

Input: nums = [-3,4,3,2], pivot = 2
Output: [-3,2,4,3]
Explanation:
The element -3 is less than the pivot so it is on the left side of the array.
The elements 4 and 3 are greater than the pivot so they are on the right side of the array.
The relative ordering of the elements less than and greater than pivot is also maintained. [-3] and [4, 3] are the respective orderings.


Constraints:

• 1 <= nums.length <= 105
• -106 <= nums[i] <= 106
• pivot equals to an element of nums.

Similar Questions:

## Solution 1. Two Pointers

This problem looks similar to the partitioning process in Quick Sort, but since the efficient partition algorithms (Lomuto’s and Hoare’s) are not stable, you can’t use it here.

• // OJ: https://leetcode.com/problems/partition-array-according-to-given-pivot/
// Time: O(N)
// Space: O(N)
class Solution {
public:
vector<int> pivotArray(vector<int>& A, int pivot) {
vector<int> gt; // numbers greater than pivot
int j = 0, cnt = 0; // j is the write pointer. cnt is the count of numbers equal to pivot
for (int i = 0; i < A.size(); ++i) {
if (A[i] < pivot) A[j++] = A[i];
else if (A[i] == pivot) ++cnt;
else gt.push_back(A[i]);
}
while (cnt--) A[j++] = pivot;
for (int i = 0; i < gt.size(); ++i) A[j++] = gt[i];
return A;
}
};

• class Solution:
def pivotArray(self, nums: List[int], pivot: int) -> List[int]:
a, b, c = [], [], []
for x in nums:
if x < pivot:
a.append(x)
elif x == pivot:
b.append(x)
else:
c.append(x)
return a + b + c

############

# 2161. Partition Array According to Given Pivot
# https://leetcode.com/problems/partition-array-according-to-given-pivot/

class Solution:
def pivotArray(self, nums: List[int], pivot: int) -> List[int]:
left = []
right = []
count = 0

for x in nums:
if x < pivot:
left.append(x)
elif x > pivot:
right.append(x)
else:
count += 1

return left + [pivot] * count + right


• class Solution {
public int[] pivotArray(int[] nums, int pivot) {
int n = nums.length;
int[] ans = new int[n];
int k = 0;
for (int x : nums) {
if (x < pivot) {
ans[k++] = x;
}
}
for (int x : nums) {
if (x == pivot) {
ans[k++] = x;
}
}
for (int x : nums) {
if (x > pivot) {
ans[k++] = x;
}
}
return ans;
}
}

• func pivotArray(nums []int, pivot int) []int {
var ans []int
for _, x := range nums {
if x < pivot {
ans = append(ans, x)
}
}
for _, x := range nums {
if x == pivot {
ans = append(ans, x)
}
}
for _, x := range nums {
if x > pivot {
ans = append(ans, x)
}
}
return ans
}


## Discuss

https://leetcode.com/problems/partition-array-according-to-given-pivot/discuss/1746999

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