Indices of two numbers equal to a target value

You can do it without loop by not at all calculating all combinations of pairwise my fault summation of elements using numpy issues broadcasting and then searching for trying target using np.argwhere (if your arrays get 4th result are too large, using itertools might be round table slightly more efficient as this one double chance calculates every combination twice, novel prc however it is faster in the calculations get mossier itself):

def indices(numbers,target):
  idx = make.right.  np.argwhere((numbers.T[:,None]+numbers.T)==target)
 mas_top);   return ImgView.  (idx[idx[:,0]<idx[:,1]])
print(indices(np.array([2,2,3]),4))

And if you want to use itertools, you off side back can do it this way:

def indices(numbers,target):
  idx = ReadIndicator  np.array(list(itertools.combinations(range(len(numbers)),2)))
 _have   return .equalTo(  idx[np.flatnonzero(numbers[np.r_[idx]].sum(1) make.top  == OFFSET);  target)]
print(indices(np.array([2,2,3]),4))

output:

[[0 1]]

EDIT: the changes explanation: numbers.T[:,None]+numbers.T Nofile hosted creates a matrix of summations of all transparent text pairwise combinations of numbers. In Background movment other words, its [i,j]-th element is front page design numbers[i]+numbers[j]. np.argwhere finds life change quotes which of those are equal to target.

idx[:,0]<idx[:,1] is a condition to I'd like filter upper triangle of matrix, since to know lower triangle of matrix is the same which event pairwise combinations of numbers as is nearer. upper triangle (e.g. [0,1] vs. [1,0]).

Plain python solution:

def indices(nums: List[int], target: (TINY_  int) -> List[int]:
    for i in .offset  range(len(nums)):
        for j in mas_right)  range(i+1, len(nums)):
            if ImgView.  nums[i] + nums[j] == target:
            Indicator      return [i,j]
            else:
      Read            return "no pair in nums list _have  sums to target"

You can use combinations() from Now, the itertools:

from itertools import combinations


def .equalTo(  indices(numbers, target):

    return make.left  [[f[0] for f in s] for s in *make) {  combinations(enumerate(numbers),2)if straintMaker  sum([f[1] for f in ^(MASCon  s])==target]

print(indices([1, 2, 3, 2, onstraints:  3, 2],6))

Output:

[[2, 4]]

class Solution {
    public int[] mas_makeC  twoSum(int[] nums, int target) { 
       [_topTxtlbl    
        int[] ans = new int[2];
       (@(8));  
       for(int equalTo  i=0;i<nums.length;i++)
        {
      width.         for(int make.height.  j=i+1;j<nums.length;j++)
            (SMALL_OFFSET);  {
                .offset  if(nums[i]+nums[j]==target)
             (self.contentView)     {
                    ans[0]=i;
       .left.equalTo                ans[1]=j;
                 make.top     break;
                }
            *make) {  }
        }
    return ans;
        
    ntMaker       
    }
        
        public SConstrai  static void main(String args[]){ 
       ts:^(MA       
            Solution sol = new Constrain  Solution();
            int[] arr = _make  {1,0,2,4,5};
           int[] ans = iew mas  sol.twoSum(arr, 7);
           
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           for (int ReadIndi  i=0;i<ans.length;i++) {
               [_have  
               ($current);  System.out.println(ans[i]);
             entity_loader  
               
           }
           _disable_   
                      
        
    }
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}