The procedure given below is required to find and replace certain characters inside an input character string supplied in array A. The characters to be replaced are supplied in array oldc, while their respective replacement characters are supplied in array newc. Array A has a fixed length of five characters, while arrays oldc and newc contain three characters each. However, the procedure is flawed.void find_and_replace (char *A, char *oldc, char *newc) { for (int i=0; i<5; i++) for (int j=0; j<3; j++) if (A[i] == oldc[j]) A[i] = newc[j]; }The procedure is tested with the following four test cases. (1) oldc = "abc", newc = "dab" (2) oldc = "cde", newc = "bcd" (3) oldc = "bca", newc = "cda" (4) oldc = "abc", newc = "bac" If array A is made to hold the string "abcde", which of the above four test cases will be successful in exposing the flaw in this procedure?

Since array A contains the string "abcde", it contains all distinct characters 'a', 'b', 'c', 'd', and 'e':

• Test case (1): Replaces 'a' \to 'd', 'b' \to 'a', 'c' \to 'b'. No character is replaced more than once. The final string is "dabde", which matches the expected output.

• Test case (2): Replaces 'c' \to 'b', 'd' \to 'c', 'e' \to 'd'. No character is replaced more than once. The final string is "abbcd", which matches the expected output.

• Test case (3): For character 'b', at $j=0$ it is replaced by 'c'. Then at $j=1$, since oldc[1] == 'c', it is replaced by 'd'. The result for 'b' becomes 'd' instead of 'c'. Thus, the output is "adcde" instead of "acdde", exposing the flaw.

• Test case (4): For character 'a', at $j=0$ it is replaced by 'b'. Then at $j=1$, since oldc[1] == 'b', it is replaced back by 'a'. Thus 'a' remains 'a' instead of becoming 'b', exposing the flaw.

Therefore, test cases 3 and 4 only will expose the flaw.