**Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target**

```
Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].
```

```
Example 2:
Input: nums = [3,2,4], target = 6
Output: [1,2]
Example 3:
```

```
Input: nums = [3,3], target = 6
Output: [0,1]
```

**Verify constraints**

Are all the numbers are positive or there be negative

```
[1,3,7,9,12] or [-1,-3,-7,-9,-12]
```

Are there duplicates no in array

```
[1,3,7,9,12] or [1,1,3,7,7,9]
```

will there be always solution available

target-25

```
[1,3,7,9,12] []
```

Solution:

**Writing Brute Force solution**

```
[1,3,7,9,2] t=11 [3,4]
```

```
[1,3,7,9,2] t=25 null
```

```
[] t=1 null
```

```
[5] t=5 null
```

```
[1,6] t=7 [0,1]
```

**Figure out solution without code**

```
0 1 2 3 4
[1,3,7,9,2] t=11 [3,4]
p1 p2
no to find= target-p[i]
11-1=10
then move array to find which element contain 10
[1,3,7,9,2]
p1 p2
[1,3,7,9,2]
p1 p2
[1,3,7,9,2]
p1 p2
```

```
0 1 2 3 4
[1,3,7,9,2] t=11 [3,4]
p1 p2
no to find= target-p[i]
11-3=17
then move array to find which element contain 7
[1,3,7,9,2]
p1 p2
[1,3,7,9,2]
p1 p2
[1,3,7,9,2]
p1 p2
```

**similary continue logic**

**First Methods**

```
const find two sum= function(num,target){
for (int i = 0; i < nums.length; i++) {
const no to find=target-nums[i];
for(int j = i+1 ; j< nums.length ; j++)
{
if(no to find == nums[j])
{
return [i,j];
}
}
}
}
```

**Another Methods**

```
public static int[] sum(int[] nums , int target)
{
int[] ans = new int[2];
if(nums.length < 2)
{
return null;
}
for (int i = 0; i < nums.length; i++) {
for(int j = i+1 ; j< nums.length ; j++)
{
if(nums[i] + nums[j] == target)
{
ans[0] = i;
ans[1] = j;
return ans;
}
}
}
return null;
}
```

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