# *Java* easy to understand iterative solution (0ms)

• The idea is very straightforward: First do a normal binary search and see if target is found.

If target is not found, return. Otherwise, find the lower bound and upper bound. To save time, one trick is that when we try to find the lower and upper bounds, we only need to scan the portion with updated `lo` and updated `hi`, rather than `0` and `nums.length-1`. This is because after the first binary search, we already know that `nums[lo]<=target` and `nums[hi]>=target`.

``````public int[] searchRange(int[] nums, int target) {
int lo=0, hi=nums.length-1, mid=0;
while(lo<=hi) {
mid = lo + ((hi-lo)>>1);
if(nums[mid]==target) break;
else if(nums[mid]>target) hi = mid-1;
else lo = mid+1;
}

if(lo>hi) return new int[]{-1,-1}; // never breaks the above while loop ==> target not found

int leftHi = mid;
while(lo<=leftHi) { // to locate lower bound for target
int midLo = (lo+leftHi)>>1;
if(nums[midLo]==target) leftHi = midLo-1;
else lo = midLo+1;
}

int rightLo = mid;
while(rightLo<=hi) { // to locate upper bound for target
int midHi = (rightLo+hi)>>1;
if(nums[midHi]==target) rightLo = midHi+1;
else hi = midHi-1;
}

return new int[]{lo, hi};
}``````

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