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All Disappeared Numbers

Certification: Basic Level Accuracy: 100% Submissions: 2 Points: 5

Write a Java program to find all the numbers that have disappeared in an array. Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once. Find all the elements of [1, n] inclusive that do not appear in this array.

Example 1
     
  • Input: nums = [4, 3, 2, 7, 8, 2, 3, 1]
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  • Output: [5, 6]
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  • Explanation: 
       
    • Step 1: The array size is 8, so the numbers should range from 1 to 8.
      •  
    • Step 2: For each number in the array, mark its presence by negating the value at index (abs(num)-1).
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    • Step 3: After marking, iterate through the array again. If a value is positive, its index+1 is a missing number.
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    • Step 4: In this case, values at indices 4 and 5 remain positive, so 5 and 6 are missing.
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    •
Example 2
     
  • Input: nums = [1, 1]
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  • Output: [2]
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  • Explanation: 
       
    • Step 1: The array size is 2, so the numbers should range from 1 to 2.
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    • Step 2: For each number in the array, mark its presence by negating the value at index (abs(num)-1).
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    • Step 3: The value at index 0 is negated twice (due to two 1's), so it becomes positive again.
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    • Step 4: The value at index 1 remains positive, so 2 is missing from the array.
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Constraints
     
  • 1 ≤ nums.length ≤ 10^5
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  • 1 ≤ nums[i] ≤ nums.length
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  • You must solve this without using extra space (or with O(1) extra space)
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  • Time Complexity: O(n) where n is the length of the array
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  • Space Complexity: O(1) not counting the output list
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ArraysNumberPwCPhillips
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Hint
Solution

Solution Hints

    
  
  • Use the array itself to mark visited numbers
    • 
  
  • Since all numbers are positive and in the range [1, n], you can use negative marking
    • 
  
  • For each number x in the array, mark the element at index (x-1) as negative
    • 
  
  • After marking, iterate through the array again. If an element is still positive, it means its index+1 is missing
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  • This approach avoids using additional space for tracking
    •

Steps to solve by this approach:

 Step 1: Initialize an empty list to store the missing numbers.
 Step 2: Since the numbers in the array are in the range [1, n], each number can be mapped to an index (number-1).
 Step 3: Iterate through the array and for each number, mark its presence by making the value at its mapped index negative.
 Step 4: If a number appears multiple times, its mapped index would already be negative, so we only negate positive values.
 Step 5: After marking, iterate through the array again.
 Step 6: If a value at index i is still positive, it means i+1 is missing from the original array.
 Step 7: Add these missing numbers to the result list and return it.

Submitted Code :