Python program to find the string weight

Given a String, each character mapped with a weight( number), compute the total weight of the string.

Input :  test_str = 'GeeksforGeeks', 
         {"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 dictionary
test_dict = {'tough': 1, 'to': 2, 'do': 3, 'todays': 4, 'work': 5}

# printing original dictionary
print("The original dictionary : " + str(test_dict))

# Initialize prefix
test_pref = 'to'

# Using dictionary comprehension + startswith()
# Prefix key match in dictionary
res = {key: val for key, val in test_dict.items()
       if key.startswith(test_pref)}

# printing result
print("Filtered dictionary keys are : " + str(res))

Output

The original string is : GeeksforGeeks
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 dictionary
test_dict = {'tough': 1, 'to': 2, 'do': 3, 'todays': 4, 'work': 5}

# printing original dictionary
print("The original dictionary : " + str(test_dict))

# Initialize prefix
test_pref = 'to'

# Using map() + filter() + items() + startswith()
# Prefix key match in dictionary
res = dict(filter(lambda item: item[0].startswith(test_pref),
                  test_dict.items()))

# printing result
print("Filtered dictionary keys are : " + str(res))

Output

The original string is : GeeksforGeeks
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 string
test_str = 'GeeksforGeeks'

# printing original string
print("The original string is : " + str(test_str))

# initializing sum dictionary
sum_dict = {"G" : 5, "e" : 2, "k" : 10,
            "f" : 3, "s" : 15, "o" : 4, "r" : 6}

# referring dict for sum
# iteration using loop
res = 0
x=list(set(test_str))
for i in x:
    res+=(test_str.count(i)*sum_dict[i])
# printing result
print("The weighted sum : " + str(res))

Output

The original string is : GeeksforGeeks
The weighted sum : 81

Approach#4: using for loop:

1. Define a function that takes two arguments, the input string and the weight dictionary.
2. Initialize a variable “weight” to 0.
3. Iterate through each character in the string.
4. If the character is present in the weight dictionary, add its weight to the “weight” variable.
5. Return the “weight” variable.

Python3

# Python program for the above approach

# Function to calculate the string weight
def 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 Code
test_str = 'GeeksforGeeks'
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.