Master the ‘Summer of 69’ Python Challenge: 5 Solutions Explained

Introduction

The author, a Python enthusiast, shares a question raised by a community member about a list‑processing problem and provides a detailed walkthrough of multiple solutions.

Problem Statement

The task, often called “Summer of 69”, requires returning the sum of numbers in an array while ignoring any sections that start with a 6 and extend to the next 9. If no numbers are present, the result should be 0.

SUMMER OF '69: Return the sum of the numbers in the array, except ignore sections of numbers starting with a 6 and extending to the next 9 (every 6 will be followed by at least one 9). Return 0 for no numbers.
summer_69([1, 3, 5]) --> 9

summer_69([4, 5, 6, 7, 8, 9]) --> 9

summer_69([2, 1, 6, 9, 11]) --> 14

Solution 1 – Xiao Wang

def summer69(arr):
    index_of_6 = arr.index(6)
    index_of_9 = len(arr) - arr[::-1].index(9) - 1
    sum_of_arr = 0
    for i, j in enumerate(arr):
        if index_of_6 <= i <= index_of_9:
            pass
        else:
            sum_of_arr += arr[i]
    return sum_of_arr

print(summer69([6, 6, 7, 8, 9, 8, 9, 2, 4]))

This approach finds the first 6 and the last 9, skips the numbers between them, and sums the rest. It fails on the first test case because it assumes both 6 and 9 are present.

Solution 2 – Yue Shen (Flag Method)

def summer_69(arr):
    total = 0
    flag = 1  # allow accumulation
    for i in arr:
        if i == 6:
            flag = 0
        elif i == 9:
            flag = 1
            continue
        total = total + i * flag
    return total

The flag variable indicates whether numbers should be added to the total.

Solution 3 – Yue Shen (While Loop)

def summer_69(arr):
    total = 0
    length = len(arr)
    left = 0
    while left < length:
        if arr[left] == 6:
            for i in range(left, length):
                left = i
                if arr[i] == 9:
                    break
        else:
            total += arr[left]
        left += 1
    return total

This method iterates with an index, skipping sections between 6 and the following 9.

Solution 4 – Chloe (Alternative)

def summer_69(arr):
    total = 0
    add = True
    for num in arr:
        while add:
            if num != 6:
                total += num
                break
            else:
                add = False
        while not add:
            if num != 9:
                break
            else:
                add = True
                break
    return total

This version toggles an “add” flag using nested while loops.

Solution 5 – Yu Liang (Recursive)

def summer_69(arr, res=0):
    if 6 in arr and 9 in arr:
        b = arr.index(6)
        e = arr.index(9)
        if b < e:
            del arr[b:e+1]
        else:
            res = res + 9
            del arr[e]
        return summer_69(arr, res)
    else:
        res += sum(arr)
        return res

print(summer_69([1, 3, 5]))
print(summer_69([4, 5, 6, 7, 8, 9]))
print(summer_69([2, 1, 6, 9, 11]))

The recursive approach removes the 6‑to‑9 segment (or handles the edge case where 9 appears before 6) and calls itself until no such segment remains.

Conclusion

The article presents five different implementations for the “Summer of 69” problem, ranging from straightforward index calculations to flag‑based loops and recursive strategies, allowing readers to compare styles and choose the one that best fits their understanding.