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gives you a string str of length n. Print the position of each element in the string so that it can form a palindrome, otherwise print the message "No palindrome" to the screen.
Palindrome is a word whose sequence of characters read from the reverse or backward direction is the same as the sequence of characters read from the forward direction, such as MADAM, racecar.
To find whether a sequence or a word is a palindrome, we usually store the reverse of the word in a separate string and compare the two, if they are the same, then the given word or sequence is a palindrome. But in this problem we have to print permutations to form words or sequences in palindromes.
Like, there is a string str = "tinni" then it can be intni or nitin so we have to return as an index starting from 1 and the resulting order can be either 2 3 1 4 5 or 3 2 1 5 4 One of the two.
The above problem requires a solution like the example given below -
Input: string str = “baa” Output: 2 1 3 Input: string str = “tinni” Output: 2 3 1 4 5
void printPalindromePos(string &str) START STEP 1: DECLARE vector<int> pos[MAX] STEP 2: DECLARE AND ASSIGN n WITH LENGTH OF str STEP 3: LOOP FOR i = 0 AND i < n AND i++ pos[str[i]].push_back(i+1) END LOOP STEP 4: SET oddCount = 0 STEP 5: DECLARE oddChar STEP 6: LOOP FOR i=0 AND i<MAX AND i++ IF pos[i].size() % 2 != 0 THEN, INCREMENT oddCount BY 1 SET oddChar AS i END IF END FOR STEP 7: IF oddCount > 1 THEN, PRINT "NO PALINDROME" STEP 8: LOOP FOR i=0 AND i<MAX AND i++ DECRLARE mid = pos[i].size()/2 LOOP FOR j=0 AND j<mid AND j++ PRINT pos[i][j] END LOOP END LOOP STEP 9: IF oddCount > 0 THEN, DECLARE AND SET last = pos[oddChar].size() - 1 PRINT pos[oddChar][last] SET pos[oddChar].pop_back(); END IF STEP 10: LOOP FOR i=MAX-1 AND i>=0 AND i-- DECLARE AND SET count = pos[i].size() LOOP FOR j=count/2 AND j<count AND j++ PRINT pos[i][j] STOP
#include <bits/stdc++.h> using namespace std; // Giving the maximum characters const int MAX = 256; void printPalindromePos(string &str){ //Inserting all positions of characters in the given string. vector<int> pos[MAX]; int n = str.length(); for (int i = 0; i < n; i++) pos[str[i]].push_back(i+1); /* find the number of odd elements.Takes O(n) */ int oddCount = 0; char oddChar; for (int i=0; i<MAX; i++) { if (pos[i].size() % 2 != 0) { oddCount++; oddChar = i; } } /* Palindrome can't contain more than 1 odd characters */ if (oddCount > 1) cout << "NO PALINDROME"; /* Print positions in first half of palindrome */ for (int i=0; i<MAX; i++){ int mid = pos[i].size()/2; for (int j=0; j<mid; j++) cout << pos[i][j] << " "; } // Consider one instance odd character if (oddCount > 0){ int last = pos[oddChar].size() - 1; cout << pos[oddChar][last] << " "; pos[oddChar].pop_back(); } /* Print positions in second half of palindrome */ for (int i=MAX-1; i>=0; i--){ int count = pos[i].size(); for (int j=count/2; j<count; j++) cout << pos[i][j] << " "; } } int main(){ string s = "tinni"; printPalindromePos(s); return 0; }
If we run the above program then it will generate the following output-
2 3 1 4 5
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