Duplicate In Array

A simple method is to traverse through the array ARR to find the frequency of each number in the given array, and we will check if the frequency of the number is more than 1.

Therefore, our approach will be to iterate currentNumber from 1 to N - 1. In each iteration, we will traverse through the array ARR to find the frequency of currentNumber in the array. We will check if the frequency is more than 1, then there is a duplicate of the number currentNumber in the array ARR. In the end, we will return the duplicate integer value present in the array. 

Algorithm:
 

• We will initialize duplicate as 0. The variable duplicate stores the duplicate element in the array.
• Iterate currentNumber from 1 to N - 1.
  • We will set count as 0. The variable count stores the count of currentNumber in the array ARR.
  • Iterate index from 0 to N - 1.
    • We will check if ARR[index] is equal to currentNumber,
      • We will Increment count by 1.
  • We will check if count is more than 1,
    • Update the value of duplicate with currentNumber.
• Return the variable duplicate.

The idea is to observe the fact that all array elements contain a value between 1 to N - 1. Our approach will be to construct an array to store the frequency of each element in the given array. 

We will construct array frequency, which will store the frequency of each element in the array. 

1. We will iterate index from 0 to N - 1, and we will increment the count of ARR[index] in the array frequency. So, we will increment frequency[ARR[index]] by 1.
2. We will iterate currentNumber from 1 to N - 1, and we will check if frequency[currentNumber] is more than 1, then there is a duplicate of the number currentNumber in the array ARR.

In the end, we will return the duplicate integer value present in the array. 

Algorithm:

• We will initialize duplicate as 0. The variable duplicate stores the duplicate element in the array.
• Create an array frequency of size N, which will store the count of each element present in the array ARR.
• Iterate index from 0 to N - 1.
  • We will set the variable currentNumber as ARR[index].
  • Increment frequency[currentNumber] by 1.
• Iterate currentNumber from 1 to N - 1.
  • We will check if frequency[currentNumber] is more than 1.
    • Update the value of duplicate with currentNumber.
• Return the variable duplicate.

The idea is to maintain two pointers, fast and slow. The pointer slow goes forward one step, and the pointer fast goes forward two steps each time. The two pointers meet in a cycle when the pointer fast becomes equal to the pointer slow, and then the duplicate number must be the entry point of the cycle.   
 

1. Our approach will be to find the cycle. We will set slow as ARR[0] and fast as    ARR[ ARR[0] ]. We will iterate till fast is not equal to small. In each iteration, we will move the pointer slow one step forward, and the pointer fast two steps forward. So, we will update slow with ARR[slow] and fast with ARR[ ARR[fast] ].
2. After we have found the cycle, we will find the entry pointer of the cycle. We will set fast as 0. We will iterate till fast is not equal to small, and in each iteration, we will update slow with ARR[slow] and fast with ARR[fast].

In the end, the variable slow will contain the duplicate element present in the array. So, we will return the variable slow.

Algorithm:
 

• We will set slow as ARR[0] and fast as ARR[ ARR[0] ].The variable slow and fast are the two-pointer that we are using to find the duplicate element in the array.
• Iterate till fast is not equal to slow.
  • Update slow with ARR[slow]. We are moving the pointer slow one step forward. 
  • Update fast with ARR[ ARR[fast] ]. We are moving the pointer fast two steps forward. 
• Assign fast as 0.
• Iterate till fast is not equal to slow.
  • Update slow with ARR[slow]. 
  • Update fast with ARR[fast].
• The variable slow will contain the duplicate element. So, we will return the variable slow.

The idea is to observe the fact that the XOR of two same numbers gives 0. The given array contains values from 1 to N - 1 once and one duplicate value. Using this idea, we can find the duplicate element val in the array ARR in the following way given below:

duplicate= (1 ^ 1) ^ (2 ^ 2) ^ (3 ^ 3)....((N - 1) ^ (N - 1)) ^ val

The XOR value of each number with itself gives 0, and the remaining value will be the duplicate value val that is present in the array ARR.

Our approach will be to initialize the variable duplicate, which will store the duplicate integer value present in the array. We will set duplicate as 0.

1. We will iterate currentNumber from 1 to N - 1, and in each iteration, we will update duplicate with XOR of duplicate and currentNumber.
2. We will iterate index from 0 to N-1, and we will update duplicate with XOR of duplicate and ARR[index].

In the end, the variable duplicate will contain the duplicate integer value present in the array ARR, and we will return the variable duplicate.

Algorithm:

• We will initialize duplicate as 0. The variable duplicate stores the duplicate element in the array.
• Iterate currentNumber from 1 to N - 1.
  • Update duplicate with XOR of duplicate and currentNumber. 
• Iterate index from 0 to N - 1.
  • Update duplicate with XOR of duplicate and ARR[index].
• Return the variable duplicate.