Approach

I approached this problem by using the two pointers solution. I used a left and right pointer, indicated by the i - left and j - right. This approach allowed me to swap the zeros with the next occurrence of a right number.

This means that when this code runs it first attempts to store a index where nums[i] == 0, and find the next non zero number directly to the right of i index. Then a swap is performed, leading to the zeros being moved to the right. This also means that j will not find an occurrence of a non zero number, and if it does they will be further moved to the RHS.

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Complexity

Time complexity: O(N)

Space complexity: O(1)

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Code

class Solution {
public:
    void moveZeroes(vector<int>& nums) {
        int i = 0, j = 0 ;
        for(; j < nums.size(); j++){
           if(nums[i] == 0 && nums[j] != 0){ 
                std::swap(nums[i], nums[j]);
                i++;
           }
           else if(nums[i] != 0){
            i++;
           }
        }
    }
};

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