19. Find the Duplicate Number

The problem can be found at the following link:

My Approach

Initialize a variable len to store the length of the input vector nums (i.e., the number of elements in the vector).
Iterate through each element num in the nums vector using a range-based for loop:
a. Calculate the absolute value of num and store it in the variable idx. This is done to ensure that we can use idx as an index to access elements in the nums vector without negative indices.
b. Check if the element at the idx position in the nums vector is already negative. If it is negative, this means that we have encountered a duplicate value because we've marked it as negative previously. In this case, return idx as the duplicate number.
c. If the element at the idx position is not negative, mark it as negative by multiplying it by -1. This is done to indicate that we have visited this element during the iteration.
After the loop, print the modified nums vector to see the changes made during the iteration (this step is for debugging purposes).
If no duplicate is found in the nums vector during the iteration, return len as a signal that no duplicates were found. However, this line of code might be an error, as it should return a duplicate number when found. You may need to modify this part of the code depending on your requirements.

Please note that the algorithm uses a technique called "negation marking" to identify duplicates. It assumes that the input vector nums contains integers in the range from 1 to len - 1 and that there is exactly one duplicate number in the vector. If these assumptions are not met, the algorithm may not work correctly.

Time and Auxiliary Space Complexity

Time Complexity: O(n)
Auxiliary Space Complexity: O(1)

Code (C++)


class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        int len = nums.size();

        for (int num : nums) {
            int idx = abs(num);
            if (nums[idx] < 0) {
                return idx;
            }
            nums[idx] = -nums[idx];
        }
        for(int num: nums) cout<<num<<" ";
        return len;
    }
};

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My Approach

Initialize a variable len to store the length of the input vector nums (i.e., the number of elements in the vector).

Iterate through each element num in the nums vector using a range-based for loop:

a. Calculate the absolute value of num and store it in the variable idx. This is done to ensure that we can use idx as an index to access elements in the nums vector without negative indices.

b. Check if the element at the idx position in the nums vector is already negative. If it is negative, this means that we have encountered a duplicate value because we've marked it as negative previously. In this case, return idx as the duplicate number.

c. If the element at the idx position is not negative, mark it as negative by multiplying it by -1. This is done to indicate that we have visited this element during the iteration.

After the loop, print the modified nums vector to see the changes made during the iteration (this step is for debugging purposes).

If no duplicate is found in the nums vector during the iteration, return len as a signal that no duplicates were found. However, this line of code might be an error, as it should return a duplicate number when found. You may need to modify this part of the code depending on your requirements.

Code (C++)


class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        int len = nums.size();

        for (int num : nums) {
            int idx = abs(num);
            if (nums[idx] < 0) {
                return idx;
            }
            nums[idx] = -nums[idx];
        }
        for(int num: nums) cout<<num<<" ";
        return len;
    }
};