Q.Write a function search_replace() in Python which accepts a list L of numbers and a number to be searched. If the number exists, it is replaced by 0 and if the number does not exist, an appropriate message is displayed. Example : L = [10,20,30,10,40] Number to be searched = 10 List after replacement : L = [0,20,30,0,40]

Here’s how you’d write the function in Python, keeping it clean and exam-ready.

The function search_replace() takes two arguments: a list L and a number n to search for. It uses a linear search — that is, it checks each element of the list one by one. If the current element equals n, it replaces that element with 0. If no match is found after checking the entire list, it prints a message saying the number was not found. The function modifies the original list directly, so no new list is returned.

def search_replace(L, n):
    found = False
    for i in range(len(L)):
        if L[i] == n:
            L[i] = 0
            found = True
    if not found:
        print("Number not found in the list.")
``` …

Let’s think about what this problem is really asking. You have a list — say [10, 20, 30, 10, 40] — and you want to check whether a particular number (like 10) appears anywhere inside it. If it does, you replace every occurrence of that number with 0. If it doesn’t appear at all, you simply print a message saying so.

This is a classic linear search task, but with a twist: instead of stopping at the first match, you must keep scanning the entire list to find and replace all matches. The NCERT Class 11 Computer Science textbook (Chapter on Lists) emphasises that lists are mutable — meaning you can change individual elements by their index. That’s the key idea here.

How do we build this function?

First, you need to know whether the number exists at all. The simplest way is to use the in operator: if n in L: returns True if n is present. But that only tells you if it exists, not where. To actually replace, you need to loop through the list using indices.

Here’s the logic step by step:

• Accept two parameters: the list L and the number to search, say num.
• Use a for loop with range(len(L)) so you can access each index.
• At each index, check if L[i] == num. If yes, set L[i] = 0.
• Keep a flag (like a boolean variable found) to track whether any replacement happened.
• After the loop, if the flag is still False, print a message like "Number not found".

Why not use list.index()?

The index() method only returns the first occurrence. You’d have to call it repeatedly in a loop, which is messy. A simple for loop over indices is cleaner and matches the NCERT approach for traversing lists.

What about the message when the number is not found?

The problem says “an appropriate message is displayed.” That means you print something like "10 not found in list" — but only if no replacement happened. So the flag is essential. …