**Problem statement**

Consider an array having N elements. Rearrange the array such that A[i] = i and if i is not present, replace it with -1 at that place. As not all elements in between 0 to N-1 will be present.

**Naive Approach
**Here we have used two for loops. One for traversing through numbers O to N-1 and the other for traversing through the array.

/*C code to Rearrange an array such that arr[i] = i*/ #include<stdio.h> void reArrangeArray(int arr[], int n) { int i, j, temp; for (i = 0; i < n; i++) { for (j = 0; j < n; j++) { if (arr[j] == i) { temp = arr[j]; arr[j] = arr[i]; arr[i] = temp; break; } } } for (i = 0; i < n; i++) { if (arr[i] != i) { arr[i] = -1; } } printf("\nArray after rearranging:\n"); for (i = 0; i < n; i++) { printf("%d\t",arr[i]); } } int main() { int n, arr[] = { -1, -1, 5, 1, 8, 3, 2, -1, 4, -1 }; n = sizeof(arr) / sizeof(arr[0]); printf("Array is:\n"); for (int i = 0; i < n; i++) { printf("%d\t",arr[i]); } reArrangeArray(arr, n); }

**Output**

gcc arri.c $ ./a.out Array is: -1 -1 5 1 8 3 2 -1 4 -1 Array after rearranging: -1 1 2 3 4 5 -1 -1 8 -1 $ |

Time complexity: O(n) |

** ****also see**

C Programming language |
Go Programming language |

Linked List |
Array |

Simplification |
Queue |

DBMS |
Reasoning |

Aptitude |
HTML |