# Python – Dictionary Keys whose Values summation equals K

Given a dictionary and a value K, extract keys whose summation of values equals K.

Input : {“Gfg” : 3, “is” : 5, “Best” : 9, “for” : 8, “Geeks” : 10}, K = 17
Output : [‘Best’, ‘for’]
Explanation : 9 + 8 = 17, hence those keys are extracted.

Input : {“Gfg” : 3, “is” : 5, “Best” : 9, “for” : 8, “Geeks” : 10}, K = 19
Output : [‘Best’, ‘Geeks’]
Explanation : 9 + 10 = 19, hence those keys are extracted.

Method #1 : Using loop

This is one of the ways in which this task can be performed. In this, we iterate for all the keys, and next keys in inner list, and keep on checking values summation. If its equal to K. The keys are stored.

## Python3

 `# Python3 code to demonstrate working of  ` `# Dictionary Keys whose Values summation equals K  ` `# Using loop ` ` `  `# initializing dictionary ` `test_dict ``=` `{``"Gfg"` `: ``3``, ``"is"` `: ``5``, ``"Best"` `: ``9``, ``"for"` `: ``8``, ``"Geeks"` `: ``10``} ` ` `  `# printing original dictionary ` `print``(``"The original dictionary is : "` `+` `str``(test_dict)) ` ` `  `# initializing K  ` `K ``=` `14` ` `  `# extracting keys and values ` `keys ``=` `list``(test_dict.keys()) ` `values ``=` `list``(test_dict.values()) ` ` `  `res ``=` `None` `for` `i ``in` `range``(``len``(keys)): ` `    ``for` `j ``in` `range``(i ``+` `1``, ``len``(keys)): ` ` `  `        ``# checking if values equals K ` `        ``if` `values[i] ``+` `values[j] ``=``=` `K: ` `            ``res ``=` `[keys[i], keys[j]] ` ` `  `# printing result  ` `print``(``"Keys whose sum equals K : "` `+` `str``(res))  `

Output

```The original dictionary is : {'Gfg': 3, 'is': 5, 'Best': 9, 'for': 8, 'Geeks': 10}
Keys whose sum equals K : ['is', 'Best']
```

Method #2 : Using list comprehension

This is yet another way in which this task can be performed. In this, we perform task similar to above method but in shorthand manner using list comprehension.

## Python3

 `# Python3 code to demonstrate working of  ` `# Dictionary Keys whose Values summation equals K  ` `# Using list comprehension  ` ` `  `# initializing dictionary ` `test_dict ``=` `{``"Gfg"` `: ``3``, ``"is"` `: ``5``, ``"Best"` `: ``9``, ``"for"` `: ``8``, ``"Geeks"` `: ``10``} ` ` `  `# printing original dictionary ` `print``(``"The original dictionary is : "` `+` `str``(test_dict)) ` ` `  `# initializing K  ` `K ``=` `14` ` `  `# extracting keys and values ` `keys ``=` `list``(test_dict.keys()) ` `values ``=` `list``(test_dict.values()) ` ` `  `# checking for keys in one liner ` `res ``=` `[[keys[i], keys[j]]  ``for` `i ``in` `range``(``len``(keys)) ``for` `j ``in` `range``(i ``+` `1``, ``len``(keys)) ``if` `values[i] ``+` `values[j] ``=``=` `K] ` ` `  `# printing result  ` `print``(``"Keys whose sum equals K : "` `+` `str``(res))  `

Output

```The original dictionary is : {'Gfg': 3, 'is': 5, 'Best': 9, 'for': 8, 'Geeks': 10}
Keys whose sum equals K : [['is', 'Best']]
```

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