{"post":"

C++ program to find the first repeating element in an array of integers

Example:

Input: {5, 15, 20, 5, 6, 10, 15, 10}

Output: 5

# Algorithm

Take an array as input and store its size in a variable n.
Start a loop from i = 0 to i < n, which will select each element from the array.
Inside the loop, start a second loop from j = i + 1 to j < n to traverse ahead and check for duplicates.
If a duplicate is found (i.e., array[i] is equal to array[j]), print the first repeating integer and return 0.
Else, print \"No integer repeated\".

Code:

#include <bits/stdc++.h>\nusing namespace std;\n\nint main() {\n   int array[100], n, i;\n   cout << \"Enter number of elements: \";\n   cin >> n;\n   cout << \"\\nEnter elements: \";\n   for (i = 0; i < n; i++)\n      cin >> array[i];\n\n   cout << \"Original array: \";\n\n   for (int i = 0; i < n; i++)\n      cout << array[i] << \" \";\n\n   // selecting an element\n\n   for (int i = 0; i < n; i++)\n      // traversing to check repetition\n      for (int j = i + 1; j < n; j++)\n         if (array[i] == array[j]) {\n            cout << \"\\nFirst repeating integer is \" << array[i];\n            return 0;\n         }\n\n   cout << \"No integer repeated\\n\";\n   return 0;\n}\n\n

Output

Enter number of elements: 5\n\nEnter elements: 1 2 3 4 5\nOriginal array: 1 2 3 4 5\nNo integer repeated\n\n

Time Complexity: O(N2)

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