First missing positive integer in linear time and constant space












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This problem was asked by Stripe.



Given an array of integers, find the first missing positive integer in
linear time and constant space. In other words, find the lowest
positive integer that does not exist in the array. The array can
contain duplicates and negative numbers as well.



For example, the input [3, 4, -1, 1] should give 2. The input [1, 2,
0] should give 3.



You can modify the input array in-place.




class DailyCodingProblem5 {

public static void main(String args) {
int arr = { 3, 4, -1, 1 };
int res = solution(arr);
System.out.println(res);

int arr2 = { 1, 2, 0 };
res = solution(arr2);
System.out.println(res);
}

private static int solution(int arr) {
int n = arr.length;
int i = 0;
for (i = 0; i < n; i++) {
int val = arr[i];

if (val <= 0 || val > n)
continue;
while (val != arr[val - 1]) {
int nextval = arr[val - 1];
arr[val - 1] = val;
val = nextval;
if (val <= 0 || val > n) {
break;
}
}
}
for (i = 0; i < n; i++) {
if (arr[i] != i + 1) {
return i + 1;
}
}
return i;

}
}


How can i improve the above solution? Is there any improvements i can have in my code ?









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    0












    $begingroup$



    This problem was asked by Stripe.



    Given an array of integers, find the first missing positive integer in
    linear time and constant space. In other words, find the lowest
    positive integer that does not exist in the array. The array can
    contain duplicates and negative numbers as well.



    For example, the input [3, 4, -1, 1] should give 2. The input [1, 2,
    0] should give 3.



    You can modify the input array in-place.




    class DailyCodingProblem5 {

    public static void main(String args) {
    int arr = { 3, 4, -1, 1 };
    int res = solution(arr);
    System.out.println(res);

    int arr2 = { 1, 2, 0 };
    res = solution(arr2);
    System.out.println(res);
    }

    private static int solution(int arr) {
    int n = arr.length;
    int i = 0;
    for (i = 0; i < n; i++) {
    int val = arr[i];

    if (val <= 0 || val > n)
    continue;
    while (val != arr[val - 1]) {
    int nextval = arr[val - 1];
    arr[val - 1] = val;
    val = nextval;
    if (val <= 0 || val > n) {
    break;
    }
    }
    }
    for (i = 0; i < n; i++) {
    if (arr[i] != i + 1) {
    return i + 1;
    }
    }
    return i;

    }
    }


    How can i improve the above solution? Is there any improvements i can have in my code ?









    share









    $endgroup$















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      0








      0





      $begingroup$



      This problem was asked by Stripe.



      Given an array of integers, find the first missing positive integer in
      linear time and constant space. In other words, find the lowest
      positive integer that does not exist in the array. The array can
      contain duplicates and negative numbers as well.



      For example, the input [3, 4, -1, 1] should give 2. The input [1, 2,
      0] should give 3.



      You can modify the input array in-place.




      class DailyCodingProblem5 {

      public static void main(String args) {
      int arr = { 3, 4, -1, 1 };
      int res = solution(arr);
      System.out.println(res);

      int arr2 = { 1, 2, 0 };
      res = solution(arr2);
      System.out.println(res);
      }

      private static int solution(int arr) {
      int n = arr.length;
      int i = 0;
      for (i = 0; i < n; i++) {
      int val = arr[i];

      if (val <= 0 || val > n)
      continue;
      while (val != arr[val - 1]) {
      int nextval = arr[val - 1];
      arr[val - 1] = val;
      val = nextval;
      if (val <= 0 || val > n) {
      break;
      }
      }
      }
      for (i = 0; i < n; i++) {
      if (arr[i] != i + 1) {
      return i + 1;
      }
      }
      return i;

      }
      }


      How can i improve the above solution? Is there any improvements i can have in my code ?









      share









      $endgroup$





      This problem was asked by Stripe.



      Given an array of integers, find the first missing positive integer in
      linear time and constant space. In other words, find the lowest
      positive integer that does not exist in the array. The array can
      contain duplicates and negative numbers as well.



      For example, the input [3, 4, -1, 1] should give 2. The input [1, 2,
      0] should give 3.



      You can modify the input array in-place.




      class DailyCodingProblem5 {

      public static void main(String args) {
      int arr = { 3, 4, -1, 1 };
      int res = solution(arr);
      System.out.println(res);

      int arr2 = { 1, 2, 0 };
      res = solution(arr2);
      System.out.println(res);
      }

      private static int solution(int arr) {
      int n = arr.length;
      int i = 0;
      for (i = 0; i < n; i++) {
      int val = arr[i];

      if (val <= 0 || val > n)
      continue;
      while (val != arr[val - 1]) {
      int nextval = arr[val - 1];
      arr[val - 1] = val;
      val = nextval;
      if (val <= 0 || val > n) {
      break;
      }
      }
      }
      for (i = 0; i < n; i++) {
      if (arr[i] != i + 1) {
      return i + 1;
      }
      }
      return i;

      }
      }


      How can i improve the above solution? Is there any improvements i can have in my code ?







      java algorithm programming-challenge





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      Maclean PintoMaclean Pinto

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