Given a String, each character mapped with a weight( number), compute the total weight of the string.
Input : test_str = 'Lazyroar',
{"G" : 1, "e" : 2, "k" : 5, "f" : 3, "s" : 15, "o" : 4, "r" : 6}
Output : 63
Explanation : 2 (G*2) + 8(e*4) + 30(s*2) + 10(k*2) + 4(o) + 6(r) +3(f) = 63.
Input : test_str = 'Geeks', {"G" : 1, "e" : 2, "k" : 5, "s" : 15}
Output : 25
Method#1: Using loop
This is one of the ways in which this task can be performed. In this, we iterate for all the characters and sum all the weights mapped from dictionary.
Python3
# Python3 code to demonstrate working of# Prefix key match in dictionary# Using dictionary comprehension + startswith()# Initialize dictionarytest_dict = {'tough': 1, 'to': 2, 'do': 3, 'todays': 4, 'work': 5}# printing original dictionaryprint("The original dictionary : " + str(test_dict))# Initialize prefixtest_pref = 'to'# Using dictionary comprehension + startswith()# Prefix key match in dictionaryres = {key: val for key, val in test_dict.items() if key.startswith(test_pref)}# printing resultprint("Filtered dictionary keys are : " + str(res)) |
Output
The original string is : Lazyroar The weighted sum : 81
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #2: Using sum()
This is one more way in which this task can be performed. In this, we use generator expression, and sum() is used to compute the summation of individual weights.
Python3
# Python3 code to demonstrate working of# Prefix key match in dictionary# Using map() + filter() + items() + startswith()# Initialize dictionarytest_dict = {'tough': 1, 'to': 2, 'do': 3, 'todays': 4, 'work': 5}# printing original dictionaryprint("The original dictionary : " + str(test_dict))# Initialize prefixtest_pref = 'to'# Using map() + filter() + items() + startswith()# Prefix key match in dictionaryres = dict(filter(lambda item: item[0].startswith(test_pref), test_dict.items()))# printing resultprint("Filtered dictionary keys are : " + str(res)) |
Output
The original string is : Lazyroar The weighted sum : 81
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #3 : Using list(),set(),count() methods
Python3
# Python3 code to demonstrate working of# String Weight# Using loop# initializing stringtest_str = 'Lazyroar'# printing original stringprint("The original string is : " + str(test_str))# initializing sum dictionarysum_dict = {"G" : 5, "e" : 2, "k" : 10, "f" : 3, "s" : 15, "o" : 4, "r" : 6}# referring dict for sum# iteration using loopres = 0x=list(set(test_str))for i in x: res+=(test_str.count(i)*sum_dict[i])# printing resultprint("The weighted sum : " + str(res)) |
Output
The original string is : Lazyroar The weighted sum : 81
Approach#4: using for loop:
- Define a function that takes two arguments, the input string and the weight dictionary.
- Initialize a variable “weight” to 0.
- Iterate through each character in the string.
- If the character is present in the weight dictionary, add its weight to the “weight” variable.
- Return the “weight” variable.
Python3
# Python program for the above approach# Function to calculate the string weightdef calculate_string_weight(test_str, weight_dict): weight = 0 for char in test_str: if char in weight_dict: weight += weight_dict[char] return weight# Driver Codetest_str = 'Lazyroar'weight_dict = {"G": 1, "e": 2, "k": 5, "f": 3, "s": 15, "o": 4, "r": 6}print(calculate_string_weight(test_str, weight_dict))test_str = 'Geeks'weight_dict = {"G": 1, "e": 2, "k": 5, "s": 15}print(calculate_string_weight(test_str, weight_dict)) |
Output
63 25
Time Complexity: O(n), where n is the length of the input string.
Space Complexity: O(1), as we are not using any additional space proportional to the input size.
