i want my output to be [i, j] for which list[i] + list[j] equals to target value. for example;
nums=[2,7,11,15]
target=9
output=[0,1] as sum of num[0]+num[1]==target
i tried my code code as;
nums=[2,7,11,15]
target=9
b=len(nums)
for i,j in zip(range(b),range(b)):
if nums[i]+nums[j]==target:
print(i,j)
i want to return position of element in list whose sums equals desired value, my above code does not ouput desired value, what will be best way to solve this problem?
You want to consider all values of j for each value of i. zip() doesn't do that, it just considers aligned pairs from each list, i.e., matching values of i and j. You need a nested loop, like this:
nums=[2,7,11,15]
target=9
b=len(nums)
for i in range(b):
for j in range(b):
if nums[i]+nums[j]==target:
print(i,j)
You can simplify it a little by using enumerate() to get indexes and values together:
nums=[2,7,11,15]
target=9
for i, n in enumerate(nums):
for j, m in enumerate(nums):
if m + n == target:
print(i,j)
There are various ways you could speed it up:
You could consider only j Indexes greater than or equal to i, and then print out both i, j and j, i when you get a match.
If you know nums is sorted, you can make it a lot faster by using a while loop that counts up from the bottom and down from the top. This would report matches when it finds them and raise the bottom counter when the sum is too low or lower the top counter when the sum is too high.
If nums is not sorted, a dictionary-based solution like @deepak-tripathi's will be most efficient. Both that and the shifting-bounds solution have O(n) solution times (vs. O(n^2) for the nested loops) and the dictionary-based solution is easier to write correctly (e.g. to handle duplicate values) and can handle sorted or unsorted lists..
Take a look at what zip() does:
list(zip(range(4), range(4)))
[(0, 0), (1, 1), (2, 2), (3, 3)]
You're not iterating over all combinations, just the cases in which i = j. Try the itertools module, it provides many efficient iterators for situations like this.
import itertools
nums=[2,7,11,15]
target=9
b=len(nums)
for i,j in itertools.combinations(range(b), 2):
if nums[i]+nums[j]==target:
print(i,j)
https://docs.python.org/3/library/itertools.html#itertools.combinations
I think this problem can be solved using dictionary like this : I have printed elements you can get the index as well
d_ = {}
nums=[2,7,11,15]
val =None
for i in nums:
if d_.get(i):
d_[i] += 1
else:
d_[i] = 1
target = 17
for i in nums:
if d_.get(target-i, None):
val = (i,target-i)
break
if val:
print(nums.index(val[0]))
print(nums.index(val[1]))
d_.get() calls with something like i in d_. Also, you could use a set instead of a dict, and create the set very quickly as set(nums). Or, if you want to show indexes instead of values, you could change your current code to store a list of indexes in d_ instead of a count of matching values. Then you could retrieve and print those indexes in the second loop.
You have to use a nested for loop.
nums = [2, 7, 11, 15]
target = 9
b = len(nums)
for i in range(b - 1):
for j in range(i + 1, b):
if nums[i] + nums[j] == target:
print(i, j)
If the list is huge or the performance matters, there are a lot of optimizations you can do with this kind of problem. For example, in the above code, sort the list first, start from small numbers. If nums[i] > target, break from the outer iteration, if nums[i] + nums[j] > target, break from the inner iteration.
Increment j by nums[j+1] and zip(range(b),range(b-1))
nums=[2,7,11,15]
target=9
b=len(nums)
for i,j in zip(range(b),range(b-1)):
if nums[i]+nums[j+1]==target:
print(i,j+1)
