Find All Numbers Disappeared in an Array – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

Given an array nums of n integers where nums[i] is in the range [1, n], return an array of all the integers in the range [1, n] that do not appear in nums.

Example 1:

Input: nums = [4,3,2,7,8,2,3,1]
Output: [5,6]

Example 2:

Input: nums = [1,1]
Output: [2]

Constraints:

• n == nums.length
• 1 <= n <= 105
• 1 <= nums[i] <= n

C++ Find All Numbers Disappeared in an Array LeetCode Solution

First iteration to negate values at position whose equal to values appear in array. Second iteration to collect all position whose value is positive, which are the missing values. Complexity is O(n) Time and O(1) space.
 
class Solution {
public:
    vector<int> findDisappearedNumbers(vector<int>& nums) {
        int len = nums.size();
        for(int i=0; i<len; i++) {
            int m = abs(nums[i])-1; // index start from 0
            nums[m] = nums[m]>0 ? -nums[m] : nums[m];
        }
        vector<int> res;
        for(int i = 0; i<len; i++) {
            if(nums[i] > 0) res.push_back(i+1);
        }
        return res;
    }
};

Java Find All Numbers Disappeared in an Array LeetCode Solution

public class Solution {
    public List<Integer> findDisappearedNumbers(int[] nums) {
        for (int i = 0; i < nums.length; i++) {
            while (nums[i] != i + 1 && nums[i] != nums[nums[i] - 1]) {
                int tmp = nums[i];
                nums[i] = nums[tmp - 1];
                nums[tmp - 1] = tmp;
            }
        }
        List<Integer> res = new ArrayList<Integer>();
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] != i + 1) {
                res.add(i + 1);
            }
        }
        return res;
    }
}
 

Python 3 Find All Numbers Disappeared in an Array LeetCode Solution

class Solution:
    def findDisappearedNumbers(self, nums: List[int]) -> List[int]:
        for i in range(len(nums)):
            temp = abs(nums[i]) - 1
            if nums[temp] > 0:
                nums[temp] *= -1
        
        res = []
        for i,n in enumerate(nums):
            if n > 0:
                res.append(i+1)
        
        return res

Array-1180
String-562
Hash Table-412
Dynamic Programming-390
Math-368
Sorting-264

Greedy-257
Depth-First Search-256
Database-215
Breadth-First Search-200
Tree-195

Binary Search-191
Matrix-176
Binary Tree-160
Two Pointers-151
Bit Manipulation-140

Stack-133
Heap (Priority Queue)-117
Design-116
Graph-108
Simulation-103
Prefix Sum-96

Backtracking-92
Counting-86
Sliding Window-73
Linked List-69
Union Find-66

Ordered Set-48
Monotonic Stack-47
Recursion-43
Trie-41
Binary Search Tree-40
Divide and Conquer-40

Enumeration-39
Bitmask-37
Queue-33
Memoization-32
Topological Sort-31

Geometry-30
Segment Tree-27
Game Theory-24
Hash Function-24
Binary Indexed Tree-21

Interactive-18
Data Stream-17
String Matching-17
Rolling Hash-17
Shortest Path-16

Number Theory-16
Combinatorics-15
Randomized-12
Monotonic Queue-9
Iterator-9
Merge Sort-9

Concurrency-9
Doubly-Linked List-8
Brainteaser-8
Probability and Statistics-7
Quickselect-7

Bucket Sort-6
Suffix Array-6
Minimum Spanning Tree-5
Counting Sort-5
Shell-4

Line Sweep-4
Reservoir Sampling-4
Eulerian Circuit-3
Radix Sort-3

Strongly Connected Componen-t2
Rejection Sampling-2
Biconnected Component-1

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