Check If the Sum of Two Different Numbers in an Array Equal a Variable Number

Check if the sum of two different numbers in an array equal a variable number?

It looks like you're trying to see if there are any two numbers in the array that add up to the specified value x. However, your code just picks two numbers at random and checks if those numbers add up.

Ruby has the Array#combination method, which generates all combinations of a given length:

def contains_pair_for_sum?(arr, n)
  !!arr.uniq.combination(2).detect { |a, b| a + b == n }
end

A few things to note:

• First, we named it according to Ruby conventions: each word is separated_by_underscores. The ? on the end means that the method is a predicate method and returns a true or false value.

• Inside the method, a few things happen. Let's look at that line, piece by piece.

  • arr: We take the array that was passed in.

  • <...>.uniq: We only look at the unique elements (because the OP wants to pick two different numbers).

  • <...>.combination(2): We ask for all combinations from the array of length 2. If the array was [4, 5, 6], we'd get [[4, 5], [4, 6], [5, 6]].

  • <...>.detect { |a, b| a + b == n }: We look for the first combination that adds up to n. If we found one, that's the result of that method. Otherwise, we get nil.

  • !!<...>: Finally, we take the result we got from detect and negate it twice. The first negation produces a Boolean value (true if the value we got was nil, or false if it's anything else); the second negation produces a Boolean value that's identical to the truth value of the first negation. This is a Ruby idiom to coerce a result into being either true or false.

Let's see it in action:

array = [4, 5, 9, 7, 8]

contains_pair_for_sum?(array, 11)
# => true (because [4, 7] sums to 11)

contains_pair_for_sum?(array, 17)
# => true (because [9, 8] sums to 17)

contains_pair_for_sum?(array, 100)
# => false (no pair matched)

Java/ Sum of any two values in an array of numbers

Should I make another array to storage sum of any two elements?

If it is required to provide only the number of the pairs equal to the given number, array is not required, simple if statement allows to check the condition and increment the counter (and print the pair if needed):

if (a[i] + a[j] == x) {
    pairs++;
    System.out.println("Found a pair: " + a[i] + " + " + a[j]);
}

What kind of method use to get the sum of any two elements in the array?

Two nested loops:

for (int i = 0; i < a.length - 1; i++) {
    for (int j = i + 1; j < a.length; j++) {
//...
    }
}

How list the number of combinations?

The total number of possible pairs is calculated by the formula of decreasing progression from n - 1 to 1: S = n * (n - 1) / 2, so for n = 5 it is 5 * 4 / 2 = 10.

The number of matching pairs is simply counted in the innermost loop.

Pair of elements from a specified array whose sum equals a specific target number

that map value you're seeing is a lookup table and that twoSum method has implemented what's called Dynamic Programming

In Dynamic Programming, you store values of your computations which you can re-use later on to find the solution.

Lets investigate how it works to better understand it:

twoSum([10,20,40,50,60,70], 50)
//I removed one of the duplicate 10s to make the example simpler

In iteration 0:

value is 10. Our target number is 50. When I see the number 10 in index 0, I make a note that if I ever find a 40 (50 - 10 = 40) in this list, then I can find its pair in index 0.

So in our map, 40 points to 0.

In iteration 2:

value is 40. I look at map my map to see I previously found a pair for 40.

map[nums[x]] (which is the same as map[40]) will return 0.

That means I have a pair for 40 at index 0.
0 and 2 make a pair.

---

Does that make any sense now?

Unlike in your solution where you have 2 nested loops, you can store previously computed values. This will save you processing time, but waste more space in the memory (because the lookup table will be needing the memory)

Also since you're writing this in javascript, your map can be an object instead of an array. It'll also make debugging a lot easier ;)

Find the two numbers that will equal the desired sum

Please have a look at this simple code because I'm pretty sure this is what you need:

function detectPair(sum, array) {  for (i=0; i<array.length; i++) {    for (j=0; j<array.length; j++) {      if (i == j) continue;      else if (array[i] + array[j] === sum) return [array[i], array[j]];    }  }; return null;}

let sum = 28;let array = [3, 5, 7, 9, 4, 8, 5, 12, 4, 9, 16, 5]
console.log(detectPair(sum, array)); //return would be [12,16]

Array b are equal to the sum the digit of the array element a

There are multiple problems in your code:

• you do not test the return value of scanf() so any invalid or missing input will cause undefined behavior because the destination variables will not be updated.

• There is a new variable n defined in the while loop, so the original variable in the outer scope is never modified by scanf().

• form and sum_digit are both incorrect: you should compute the sum in sum_digit() with a loop and store all sums in a loop in form.

Here is a modified version:

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

#define MAX 100

void form(int [], int [], int);
int sum_digits(int);
void print (int [], int);

int main() {
    int n;
    
    for (;;) {
        printf("Enter n:");
        if (scanf("%d", &n) != 1)
            return 1;
        if (n >= 1 && n <= MAX)
            break;
        printf("Incorrect value: %d\n", n);
    }
    int a[MAX], b[MAX];
    printf("Enter elements of array a:\n");
    for (int i = 0; i < n; i++) {
        printf("%d. element:", i + 1);
        if (scanf("%d", &a[i]) != 1)
            return 1;
    }
    form(a, b, n);
    print(b, n);
    return 0;
}

void form(int a[MAX], int b[MAX], int n) {
    for (int i = 0; i < n; i++) {
        b[i] = sum_digits(a[i]);
    }
}

int sum_digits(int s) {
    int sum = 0;

    /* this loop handles positive and negative numbers alike */
    while (s != 0) {
        sum = sum + s % 10;
        s = s / 10;
    }
    /* return the positive sum of digits */
    return abs(sum);
}

void print(int b[MAX], int n) {
    for (int i = 0; i < n; i++) {
        printf("%d ", b[i]);
    }
    printf("\n");
}

Compare items from two arrays to see if they sum to 10

If you haven't figured out the solution, here it is:

private static string NumberOfTens(int number1, int number2)
    {
      //Store 10-pairs in a new variable
      int tenPairs = 0;

      // Need variables to define the dice - arrays starting at 1?
      int[] die1 = Enumerable.Range(1, number1).ToArray();

      int[] die2 = Enumerable.Range(1, number2).ToArray();

      //Compare each item from die1 to each value from die2 to see if they add up to 10.

      for (int i = 0; i < die1.Length; i++)
      {
        for (int j = 0; j < die2.Length; j++)
        {
          if (die1[i] + die2[j] == 10)
          {
            tenPairs++;
          }
        }
      }

      //return a string with the message: "There are X total ways to get the sum 10."
      string resultMsg = "There are " + tenPairs + " total ways to make 10.";

      return resultMsg;
    }

here is how to use it :

Console.WriteLine(NumberOfTens(5, 15));

php find sum of two numbers in a dynamic array and see if they equal a given number

Sounds like all you need is to use array_slice before using array_sum.

Something like:

function subarr_sum($array,$offset,$length) {
  $newarray = array_slice($array,$offset,$length);
  return array_sum($newarray);
}

---

Related Topics