Print all possible combinations of strings that can be made using a keypad given a number

Consider a phone keypad where every number has few english alphabet characters associated with it. When a user types some numbers on screen , print all possible permutations of string possible using the characters associated with those numbers. Example: if 1 corresponds to A B C and 2 corresponds to DEF. Then number "12" might represent any of the following : AD, AE, AF, BD, BE, BF, CD, CE, CF.
The problem is very similar to this problem.
The key steps in writing an this function are:

• Initialize a keypad array of strings with alphabets corresponding to each number.
• Now one by one choose each of the characters corresponding to the number and recurse.
• Make sure to keep a check for the digit 9 as it will have only 2 characters.

string keypad[9] = {"ABC", "DEF", "GHI", "JKL", "MNO", "PQR", "STU", "VWX", "YZ"};

void keys(char* num, char *s, int start, int end)
{
    if(start == end)
    {
        cout<<s<<endl;
        return;
    }
    
    s[start] = keypad[(num[start] - '0' - 1)][0];
    keys(num, s,start+1,end);
    s[start] = keypad[(num[start] - '0' - 1)][1];
    keys(num, s,start+1,end);
    if((num[start] - '0' - 1) < 8)
    {
        s[start] = keypad[(num[start] - '0' - 1)][2];
        keys(num, s,start+1,end);
    }
}

Function to print all permutations of a string!!

Write a function to print all permutations of a string. Characters in the string do not repeat themselves.
The problem is pretty similar to this problem.
Solution is a simple recursive approach where we call the function again and again once with each arrangement of characters. These arrangements are achieved by swapping start with every other character and calling the function again and again with different values of start.

void perm(string s, int start, int end)
{
    //If we have already reached the end of string, print it
    if(start == end)
    {
        cout<<s<<endl;
        return;
    }
    char temp;
    for(int i=start;i<end;i++)
    {
        SWAP(s[start],s[i],temp);
        perm(s,start+1,end);
        SWAP(s[i],s[start],temp);
    }
}

Unsolved!!

Here are some questions which I will try in coming days and have not solved them. Will take them of this list as I solve each one of them :).

• Given a linked list structure where every node represents a linked list and contains two pointers of its type: (i) pointer to next node in the main list. (ii) pointer to a linked list where this node is the head. Write a function to flatten the list into a single linked list.


• You are given an array of integers A[1..n] and a maximum sliding window of size w. Output an array B where B[i] is the maximum in A[i, i+w-1].


• You have an array of 0s and 1s and you want to output all the intervals (i, j) where the number of 0s and numbers of 1s are equal.


• Given a binary matrix of NxN of integers ,return only unique rows of binary arrays eg: 01001 10110 01001 11100 ans: 01001 10110 11100


• Given an array of integers like ar[]= {1,3,2,4,5,4,2}. Create another array ar_low[] such that ar_low[i] = # of elements lower than or equal to ar[i] in ar[i+1:n-1]. So the output for above array should be {0,2,1,2,2,1,0}