**ObjecÂtive:** **– **Given an Integer array. Array contains duplicates of all the numbers in array except one number . Find that number.

**Example** :

int [] A = { 2,1,3,5,5,3,2,1,6,7,7,8,8}; Output : Missing duplicate is 6

**Appraoch:**

**Naive solution**is use Hash Table ..space complexity – O(n)**Better solution****– XOR**- A^A = 0 and A^B^A = B, so if we XOR all the elements, answer will be the missing no
- If we have only one element, the missing no will be that no

**Code:**

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public class MissingDuplicate { | |

// naive solution is use Hash Table ..space complexity – O(n) | |

// better solutiion – XOR | |

// A^A = 0 and A^B^A = B, so if we XOR all the elements, answer will be the | |

// missing no | |

public int find(int[] A) { | |

int miss = A[0]; // if we have only one element, the missing no will be | |

// that no | |

for (int i = 1; i < A.length; i++) { | |

miss = miss ^ A[i]; | |

} | |

return miss; | |

} | |

public static void main(String[] args) { | |

int[] A = { 2, 1, 3, 5, 5, 3, 2, 1, 6, 7, 7, 8, 8 }; | |

MissingDuplicate i = new MissingDuplicate(); | |

System.out.println("Missing duplicate is " + i.find(A)); | |

} | |

} |

**Output**:

Missing duplicate is 6

Nice One