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2161. Partition Array According to Given Pivot

Description

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.

Solutions

Solution 1: Simulation

We can traverse the array $\textit{nums}$, sequentially finding all elements less than $\textit{pivot}$, all elements equal to $\textit{pivot}$, and all elements greater than $\textit{pivot}$, then concatenate them in the order required by the problem.

Time complexity $O(n)$, where $n$ is the length of the array $\textit{nums}$. Ignoring the space consumption of the answer array, the space complexity is $O(1)$.

Solution 2: Two pointers

This implementation follows the required operations directly. It traverses the relevant values and updates its state as each value is processed. After all required states have been considered, the maintained result is returned.

  • 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;
    }
}

  • class Solution {
public:
    vector<int> pivotArray(vector<int>& nums, int pivot) {
        vector<int> ans;
        for (int& x : nums)
            if (x < pivot) ans.push_back(x);
        for (int& x : nums)
            if (x == pivot) ans.push_back(x);
        for (int& x : nums)
            if (x > pivot) ans.push_back(x);
        return ans;
    }
};

  • 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


  • 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
}

  • function pivotArray(nums: number[], pivot: number): number[] {
    const ans: number[] = [];
    for (const x of nums) {
        if (x < pivot) {
            ans.push(x);
        }
    }
    for (const x of nums) {
        if (x === pivot) {
            ans.push(x);
        }
    }
    for (const x of nums) {
        if (x > pivot) {
            ans.push(x);
        }
    }
    return ans;
}


// Solution 2
function pivotArray(nums: number[], pivot: number): number[] {
    const n = nums.length;
    const res = Array(n).fill(pivot);

    for (let i = 0, l = 0, j = n - 1, r = n - 1; i < n; i++, j--) {
        if (nums[i] < pivot) res[l++] = nums[i];
        if (nums[j] > pivot) res[r--] = nums[j];
    }

    return res;
}


  • function pivotArray(nums: number[], pivot: number): number[] {
    const n = nums.length;
    const res = Array(n).fill(pivot);

    for (let i = 0, l = 0, j = n - 1, r = n - 1; i < n; i++, j--) {
        if (nums[i] < pivot) res[l++] = nums[i];
        if (nums[j] > pivot) res[r--] = nums[j];
    }

    return res;
}


  • function pivotArray(nums, pivot) {
    const n = nums.length;
    const res = Array(n).fill(pivot);

    for (let i = 0, l = 0, j = n - 1, r = n - 1; i < n; i++, j--) {
        if (nums[i] < pivot) res[l++] = nums[i];
        if (nums[j] > pivot) res[r--] = nums[j];
    }

    return res;
}

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