CODEWITHSUNDEEP
java
john
john

Reputation: 201

finding how many consecutive numbers in an array [java]

For example, array is like this {1,1,2,1,1,1 } and they key int is 1, the Largest number of consecutive times 1 is going to be 3. When i run my code and type the same numbers, I'm getting 4 consecutive numbers of 1. Can someone give me advice? is there a built-in method that will help me solve this?

 Scanner kbd = new Scanner (System.in);
            int count = 1;
            int largest = 0;

            System.out.println("Enter number");
            int numbers = kbd.nextInt();
            int[] numb = new int[numbers];
            System.out.println("Now enter "+numbers+" integers:");
            for(int i =0; i<numb.length;i++){
                numb[i] = kbd.nextInt();
            }
            System.out.println("Now enter the key integer: ");
            int key = kbd.nextInt();

            for (int i = 0; i<numb.length-1; i++) {
            if (numb[i] == key) {
                if (numb[i] == numb[i + 1]) {
                     count++;
                    }
            else {
                    largest = Math.max(largest, count);
                    count = 1;

                  }
           }
    }

            System.out.println("Largest number of consecutive times "+key+" was entered: "+largest);

Upvotes: 1

Views: 12556

Answers (5)

cody123
cody123

Reputation: 2080

Try with this.

        int numb[] = new int[] { 6, 3, 3, 5,5,5 ,3, 5, 15,15 };
        int key = 3; // Can be changed
        int largest = 0;
        int currCount = 0;
        for (int i = 0; i < numb.length; i++) {
            if (numb[i] == key) {
                currCount++;
            } else {
                if (currCount > largest) {
                    largest = currCount;
                }
                currCount = 0;
            }
        }
        if (currCount > largest) {
            largest = currCount;
        }

        System.out.println("Largest number of consecutive times " + key + " was entered: " + largest);

Upvotes: 0

Addis
Addis

Reputation: 2530

for (int i = 0; i<numb.length; i++) {
    if (numb[i] == key) {
        if(i == numb.length-1){
            count =1;
        }
        else if (numb[i] == numb[i+1]) {
            count++;
        }
        if (count > largest) {
            largest = count;
        }
    }
    else
        count = 1;
}

Upvotes: 0

Ravikumar
Ravikumar

Reputation: 901

Just reassign count to 0 in the if condition (In your initial code before editing).

if(largest<count){
   largest = Math.max(largest, count);
   count = 0;
}

Output

And in the current code just add a if condition checking whether largest is greater than count after the for loop.

largest = (largest > count) ? largest : count;

Or

if (count > largest){
   largest = count;
}

Upvotes: 0

Philip Tinney
Philip Tinney

Reputation: 2016

A slightly different approach.

count = 0;
for (int i : numb){
  if (i == key){
    count++;
  } else {
    count = 0;
  }
  if(largest<count) {
    largest = count;
  }
}

Upvotes: 2

Louis Wasserman
Louis Wasserman

Reputation: 198103

It doesn't look like you should have a second inner loop, it should just be

for (int i = 0; i + 1<numb.length; i++){
   if(numb[i] == numb[i + 1]){
      count++;
   }else{
      if(largest<count) {
        largest = count;
        count = 1;
   }
 }

Upvotes: 1

Related Questions