Search for a Range

Search for a Range Total Accepted: 36248  Total Submissions: 130885

Given a sorted array of integers, find the starting and ending position of a Given target value.

Your algorithm ' s runtime complexity must is in the order of O(log n).

if The target is not a found in the array, Return [-1,-1] .

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
Return [3, 4] .

MySolution:

int returnarray[2]={-1,-1};int*searchrange (int a[], int n, int target) {int startingdex = 0,endingdex = 0;    int i;    int isfoundtarget = 0;    int isthelastelement = 0;    int ihalf = (int) (N/2)-1;    /* 0 N/2 n-1 */* 0 N/2 */if (Ihalf < 0) ihalf = 0; if (target >= a[0] && target <= A[ihalf]) {for (i = 0; I <= ihalf; i++) {i F (target = = A[i] && Isfoundtarget = = 0) {/* found the target */Isfoun                Dtarget=1;                Startingdex = i;                Endingdex = Startingdex;            Continue                } else {if (target = = A[i]) {endingdex = i;            } continue; }}}/* N/2 n-1 */if (target >= a[ihalf] && target <=a[n-1]) {for (i = ihalf; I <= n-1;           i++) {if (i = = n-1)     Isthelastelement = 1;                if (target = = A[i]) {if (Isfoundtarget = = 0)/* Startingdex not found before */                    {isfoundtarget = 1;                Startingdex = i; } if (Isfoundtarget! = 0 | |                    Isthelastelement! = 0) {Endingdex = i;                Continue    }}}} if (Isfoundtarget = = 0) {Startingdex = Endingdex =-1;    } returnarray[0] = Startingdex;    RETURNARRAY[1] = Endingdex; return ReturnArray;}

Result:

Submission result:accepted

Search for a Range