Count number of occurrences of a string

Get the number of occurrences of a sub string in the given string.

As you see above image string “ab” is repeated twice in “abrakadabra”.

Procedure

Step 1: Let sourceString=”abrakadabra”, substring=”ab”

Step 2: Start from the 0^th index of the source string and search for the substring “ab”, if you do not find any sib string, then there are 0 occurrences. If you found a sub string, then start from the index (where the substring is located in the sourceString) + substring.legth() in sourceString.

Repeat this procedure until end of the string.

Find the below working application.

StringUtil.java

package com.sample.utils;

public class StringUtil {

 /**
  * It return the number of occurrences of <code>subString</code> in the
  * <code>sourceString</code>
  * 
  * @param sourceString
  * @param subString
  * @return
  */
 public static int noOfOccurrences(String sourceString, String subString) {
  if (sourceString == null || subString == null || subString.isEmpty() || sourceString.isEmpty()
    || (subString.length() > sourceString.length())) {
   return -1;
  }

  int index = 0;
  int noOfTimes = 0;
  int fromIndex = 0;

  while ((index = sourceString.indexOf(subString, fromIndex)) != -1) {
   noOfTimes++;
   fromIndex = index + subString.length();  
  }
  
  return noOfTimes;

 }
}

StringUtilTest.java

package com.sample.util;

import static org.junit.Assert.assertEquals;

import org.junit.Test;

import com.sample.utils.StringUtil;

public class StringUtilTest {

 private String sourceString = null;
 private String subString = null;
 private int noOfOccurrences = -1;

 @Test
 public void noOfOccurrences_nullSourceAndnullSubString_negative1() {
  noOfOccurrences = StringUtil.noOfOccurrences(null, null);
  assertEquals(noOfOccurrences, -1);
 }

 @Test
 public void noOfOccurrences_nullSourceAndEmptySubString_negative1() {
  noOfOccurrences = StringUtil.noOfOccurrences(null, "");
  assertEquals(noOfOccurrences, -1);
 }

 @Test
 public void noOfOccurrences_emptySourceAndNullSubString_negative1() {
  noOfOccurrences = StringUtil.noOfOccurrences("", null);
  assertEquals(noOfOccurrences, -1);
 }

 @Test
 public void noOfOccurrences_emptySourceAndEmptySubString_negative1() {
  noOfOccurrences = StringUtil.noOfOccurrences("", "");
  assertEquals(noOfOccurrences, -1);
 }

 @Test
 public void noOfOccurrences_sourceStringAndSubString_negative1() {
  sourceString = "abrakadabra";
  subString = "ab";

  noOfOccurrences = StringUtil.noOfOccurrences(sourceString, subString);
  assertEquals(noOfOccurrences, 2);
 }

}

You may like

Miscellaneous

Programming Questions

Interview Questions

Find the missing element

Find the max product value of 3 numbers

Find the first element that appears an even number of times

Find the element that appears once in an array

Convert integers into roman equivalents