Time limit exceeded on HackerRank “Username Disparity”












0












$begingroup$


Here is the hackerrank question:




For two strings A and B, we define the similarity of the strings to be the length of the longest prefix common to both strings. For example, the similarity of strings "abc" and "abd" is 2, while the similarity of strings "aaa" and "aaab" is 3.






I am getting a timeout on some of the test cases but not all of them.



function usernameDisparity(inputs) {
return inputs.map(function (input) {
return getDisparity(input);
});
}


function getDisparity(input) {
var sum = 0;
for (var i = 0; i < input.length; i++) {
sum = sum + getSimilarity(i, input);
}
return sum;
}

function getSimilarity(index, input) {
var similarity = 0;
for (var i = 0; i < input.length; i++, index++) {
if (input[index] !== input[i]) {
return similarity;
} else {
similarity = similarity + 1;
}
}
return similarity;
}









share|improve this question









New contributor




Zhen Liu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

















    0












    $begingroup$


    Here is the hackerrank question:




    For two strings A and B, we define the similarity of the strings to be the length of the longest prefix common to both strings. For example, the similarity of strings "abc" and "abd" is 2, while the similarity of strings "aaa" and "aaab" is 3.






    I am getting a timeout on some of the test cases but not all of them.



    function usernameDisparity(inputs) {
    return inputs.map(function (input) {
    return getDisparity(input);
    });
    }


    function getDisparity(input) {
    var sum = 0;
    for (var i = 0; i < input.length; i++) {
    sum = sum + getSimilarity(i, input);
    }
    return sum;
    }

    function getSimilarity(index, input) {
    var similarity = 0;
    for (var i = 0; i < input.length; i++, index++) {
    if (input[index] !== input[i]) {
    return similarity;
    } else {
    similarity = similarity + 1;
    }
    }
    return similarity;
    }









    share|improve this question









    New contributor




    Zhen Liu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      0












      0








      0





      $begingroup$


      Here is the hackerrank question:




      For two strings A and B, we define the similarity of the strings to be the length of the longest prefix common to both strings. For example, the similarity of strings "abc" and "abd" is 2, while the similarity of strings "aaa" and "aaab" is 3.






      I am getting a timeout on some of the test cases but not all of them.



      function usernameDisparity(inputs) {
      return inputs.map(function (input) {
      return getDisparity(input);
      });
      }


      function getDisparity(input) {
      var sum = 0;
      for (var i = 0; i < input.length; i++) {
      sum = sum + getSimilarity(i, input);
      }
      return sum;
      }

      function getSimilarity(index, input) {
      var similarity = 0;
      for (var i = 0; i < input.length; i++, index++) {
      if (input[index] !== input[i]) {
      return similarity;
      } else {
      similarity = similarity + 1;
      }
      }
      return similarity;
      }









      share|improve this question









      New contributor




      Zhen Liu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      Here is the hackerrank question:




      For two strings A and B, we define the similarity of the strings to be the length of the longest prefix common to both strings. For example, the similarity of strings "abc" and "abd" is 2, while the similarity of strings "aaa" and "aaab" is 3.






      I am getting a timeout on some of the test cases but not all of them.



      function usernameDisparity(inputs) {
      return inputs.map(function (input) {
      return getDisparity(input);
      });
      }


      function getDisparity(input) {
      var sum = 0;
      for (var i = 0; i < input.length; i++) {
      sum = sum + getSimilarity(i, input);
      }
      return sum;
      }

      function getSimilarity(index, input) {
      var similarity = 0;
      for (var i = 0; i < input.length; i++, index++) {
      if (input[index] !== input[i]) {
      return similarity;
      } else {
      similarity = similarity + 1;
      }
      }
      return similarity;
      }






      javascript strings programming-challenge time-limit-exceeded






      share|improve this question









      New contributor




      Zhen Liu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|improve this question









      New contributor




      Zhen Liu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|improve this question




      share|improve this question








      edited 5 mins ago









      200_success

      129k15153415




      129k15153415






      New contributor




      Zhen Liu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      asked 14 hours ago









      Zhen LiuZhen Liu

      101




      101




      New contributor




      Zhen Liu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      Zhen Liu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      Zhen Liu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






















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