I have 2 strings where I have saved fixed 20 characters and these are “A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T” and same 20 char in string 2. so I will get 400 combinations of 2 character sets like Code: AA,AB,AC,AD,AE,AF,……………AT BA,BB,BC,BD,BE,BF,…………..BT CA,CB,CC,CD,CE,CF……………CT Code: #include<stdio.h> #include<conio.h> #include<string.h> int main() { int count = 0; char sequence1[30] = "ACDEFGHILPNQYWRKMTSV"; char sequence2[30] = "ACDEFGHILPNQYWRKMTSV"; const char sequence3[30] = "AAGACDF"; /*sequence taken from user which will be of length 31*/ printf(" %s %s \n ",sequence1,sequence2); for (int i=0; i<strlen(sequence1); i++) { for(int j=0;j<strlen(sequence2);j++) { printf(" %c %c",sequence1[i],sequence2[j]); for(int k=0; k < strlen(sequence3)-1; k++) { if(sequence3[k] == sequence1[i] && sequence3[k+1] == sequence2[j]) { //cout<<"(equal)\t"; count++; } } } } printf("the number of matches are found %d ",count); getch(); } This way we will get 400 combinations (Which program I have created successfully) but then user will put the value till 31 characters witch will be treated as 3rd string for E.g. “ABCDDAAAB” now I have to check the frequency of user input in the sequence of 12,23,34,45,56,67,78,89 (2 CHAR SET) means AB,BC,CD,DD,DA,AA,AA,AB and need to show the frequency of user input OUTPUT: AB=2 BC=1 CD=1 DD=1 DA=1 AA=2
1: 2: How are the two (1 and 2) linked ?? The problem that you mentioned in 2 can be solved independently !