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# Python Program that displays the key of list value with maximum range

Given a Dictionary with keys and values that are lists, the following program displays key of the value whose range in maximum.

Range = Maximum number-Minimum number

Input : test_dict = {“Gfg” : [6, 2, 4, 1], “is” : [4, 7, 3, 3, 8], “Best” : [1, 0, 9, 3]}
Output : Best
Explanation : 9 – 0 = 9, Maximum range compared to all other list given as values

Input : test_dict = {“Gfg” : [16, 2, 4, 1], “Best” : [1, 0, 9, 3]}
Output : Gfg
Explanation : 16 – 1 = 15, Maximum range compared to all other list given as values

Method 1: Using max(), min() and loop

In this, we get max() and min() of each list and perform difference to find range. This value is then stored and max difference of all such values is computed by applying max() at the result list.

## Python3

 # initializing dictionarytest_dict = {"Gfg" : [6, 2, 4, 1], "is" : [4, 7, 3, 3, 8], "Best" : [1, 0, 9, 3]} # printing original dictionaryprint("The original dictionary is : " + str(test_dict)) max_res = 0for sub, vals in test_dict.items():         # Storing maximum of difference    max_res = max(max_res, max(vals) - min(vals))       if max_res == max(vals) - min(vals):        res = sub         # printing resultprint("The maximum element key : " + str(res))

Output

The original dictionary is : {'Gfg': [6, 2, 4, 1], 'is': [4, 7, 3, 3, 8], 'Best': [1, 0, 9, 3]}
The maximum element key : Best

Time Complexity: O(n) where n is the number of elements in the dictionary “test_dict”. The sorted and itemgetter function is used to perform the task and it takes O(n*logn) time.
Auxiliary Space: O(n) additional space of size n is created where n is the number of elements in the dictionary “test_dict”.

Method 2: Using list comprehension, max() and min()

In this, we compute the maximum range, and then extract the key which matches that difference using list comprehension.

## Python3

 # initializing dictionarytest_dict = {"Gfg" : [6, 2, 4, 1], "is" : [4, 7, 3, 3, 8], "Best" : [1, 0, 9, 3]} # printing original dictionaryprint("The original dictionary is : " + str(test_dict)) # getting max valuemax_res = max([max(vals) - min(vals) for sub, vals in test_dict.items()]) # getting key matching with maximum valueres = [sub for sub in test_dict if max(test_dict[sub]) - min(test_dict[sub]) == max_res][0] # printing resultprint("The maximum element key : " + str(res))

Output

The original dictionary is : {'Gfg': [6, 2, 4, 1], 'is': [4, 7, 3, 3, 8], 'Best': [1, 0, 9, 3]}
The maximum element key : Best

Method 3: Finding the Key with the Maximum Difference in Values in a Dictionary.

• Initialize two variables max_diff and max_key to None.
• Iterate over the items in the dictionary using a for loop.
• For each item, calculate the difference between its maximum and minimum value using max() and min() functions.
• If max_diff is None or the difference for the current item is greater than max_diff, update max_diff and max_key to the current difference and key respectively.
• After the loop, max_key will contain the key with the maximum difference.

## Python3

 # initializing dictionarytest_dict = {"Gfg" : [6, 2, 4, 1], "is" : [4, 7, 3, 3, 8], "Best" : [1, 0, 9, 3]} # printing original dictionaryprint("The original dictionary is : " + str(test_dict)) # alternative approachmax_diff = Nonemax_key = None for key, value in test_dict.items():    diff = max(value) - min(value)    if max_diff is None or diff > max_diff:        max_diff = diff        max_key = key # printing resultprint("The maximum element key : " + str(max_key))

Output

The original dictionary is : {'Gfg': [6, 2, 4, 1], 'is': [4, 7, 3, 3, 8], 'Best': [1, 0, 9, 3]}
The maximum element key : Best

Time complexity: O(n), where n is the number of items in the dictionary.
Auxiliary space: O(1), as it only uses a constant amount of space to store the max_diff and max_key variables.

Method 4: Using the built-in ‘map’ function

• Define a function that computes the difference between the maximum and minimum value of a list.
• Use the ‘map’ function to apply the function to each value of the dictionary.
• Find the index of the maximum difference using the ‘index’ method of the resulting list.
• Use the ‘list’ method to retrieve the keys of the dictionary and get the corresponding key.

## Python3

 # initializing dictionarytest_dict = {"Gfg": [6, 2, 4, 1], "is": [4, 7, 3, 3, 8], "Best": [1, 0, 9, 3]} # function to compute the difference between the maximum and minimum value of a listdef max_min_diff(lst):    return max(lst) - min(lst) # computing difference between max and min values of each list in the dictionary using map functiondiffs = list(map(max_min_diff, test_dict.values())) # finding the index of maximum differencemax_idx = diffs.index(max(diffs)) # retrieving corresponding key from dictionarymax_key = list(test_dict.keys())[max_idx] # printing resultprint("The maximum element key : " + str(max_key))

Output

The maximum element key : Best

Time complexity: O(n*m), where n is the number of keys and m is the length of the longest list in the dictionary.
Auxiliary space: O(n), for creating the list of differences.

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