S1Q3 · Swap the Last Chars in Dictionary Values¶

⚡ Quick Reference

Function: swap_last_chars_of_values(d: dict, k1, k2) -> None

Core idea: extract the last character of each value, swap them, rebuild both strings.

def swap_last_chars_of_values(d: dict, k1, k2):
    v1, v2 = d[k1], d[k2]
    d[k1] = v1[:-1] + v2[-1]
    d[k2] = v2[:-1] + v1[-1]

Key rules: - v[-1] → last character - v[:-1] → everything except the last character - Modifies the dict in-place - returns None - Save both last chars before modifying (avoid overwriting)

Problem Statement¶

Problem

Write a function swap_last_chars_of_values(d, k1, k2) that swaps the last characters of the string values at keys k1 and k2 in dictionary d, modifying it in-place.

Examples:

Input

d={"first":"apple","second":"mango","third":"banana"}, k1="first", k2="second"

Output (modified dict)

{"first": "applo", "second": "mange", "third": "banana"}

Input

d={"key1":"hello","key2":"world"}, k1="key1", k2="key2"

Output (modified dict)

{"key1": "helld", "key2": "worlo"}

Tracing both examples¶

Example 1: d["first"] = "apple", d["second"] = "mango"

v1 = "apple",  v2 = "mango"
v1[-1] = 'e',  v2[-1] = 'o'

d["first"]  = "appl" + "o" = "applo"
d["second"] = "mang" + "e" = "mange"

"third" is untouched ✓

Example 2: d["key1"] = "hello", d["key2"] = "world"

v1 = "hello",  v2 = "world"
v1[-1] = 'o',  v2[-1] = 'd'

d["key1"] = "hell" + "d" = "helld"
d["key2"] = "worl" + "o" = "worlo"

✓

Why save both values upfront?

If you updated d[k1] first and then read d[k1][-1] for the swap, you'd get the already-modified last character - not the original one. Always read both original values into local variables before writing back.

Solution approaches¶

Pythonic (clean one-liner logic)Explanatory (named variables)Using tuple unpackingUsing lambda

def swap_last_chars_of_values(d: dict, k1, k2):
    v1, v2 = d[k1], d[k2]
    d[k1] = v1[:-1] + v2[-1]
    d[k2] = v2[:-1] + v1[-1]

def swap_last_chars_of_values(d: dict, k1, k2):
    val1 = d[k1]
    val2 = d[k2]

    last1 = val1[-1]    # last char of first value
    last2 = val2[-1]    # last char of second value

    d[k1] = val1[:-1] + last2   # replace last char of val1 with last2
    d[k2] = val2[:-1] + last1   # replace last char of val2 with last1

def swap_last_chars_of_values(d: dict, k1, k2):
    # Python evaluates the right side fully before assigning
    d[k1], d[k2] = d[k1][:-1] + d[k2][-1], d[k2][:-1] + d[k1][-1]

Python evaluates the entire right-hand side before any assignment, so the original values of d[k1] and d[k2] are used in both expressions - no temporary variables needed.

def swap_last_chars_of_values(d: dict, k1, k2):
    swap = lambda a, b: (a[:-1] + b[-1], b[:-1] + a[-1])
    d[k1], d[k2] = swap(d[k1], d[k2])

Key takeaways¶

01

v[:-1] + new_char replaces the last character

v[:-1] slices everything except the last character. Concatenating a new character gives the modified string. This works for any string length since strings are immutable - you build a new one.

02

Read both values before writing either

Storing v1, v2 = d[k1], d[k2] before any modification ensures you use the original last characters for both swaps. Writing d[k1] first would corrupt the value needed for d[k2].

03

In-place modification - return None

The function modifies the dictionary directly and returns nothing. The caller sees the changes through the same dict reference. Don't return a new dict - that would leave the original unchanged.