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JavaScript: Check whether two arrays of integers of same length are similar or not

JavaScript Basic: Exercise-87 with Solution

Write a JavaScript program to check whether two arrays of integers of same length are similar or not. The arrays will be similar if one array can be obtained from another array by swapping at most one pair of elements.

Pictorial Presentation:

JavaScript: Check whether two arrays of integers of same length are similar or not.

Sample Solution:

HTML Code:

<!DOCTYPE html>
<html>
<head>
  <meta charset="utf-8">
  <meta name="viewport" content="width=device-width">
  <title>Write a JavaScript program to check whether two arrays of integers of same length are similar or not. The arrays will be similar if one array can be obtained from another array by swapping at most one pair of elements</title>
</head>
<body>

</body>
</html>

JavaScript Code:

function array_checking(arra1, arra2) {

  for(var i = 0; i < arra1.length; i++) {
    for(var j = i; j < arra1.length; j++) {
      var result = true,
        temp = arra1[i];
      arra1[i] = arra1[j];
      arra1[j] = temp;
      for(var k = 0; k < arra1.length; k++) {
        if(arra1[k] !== arra2[k]) {
          result = false;
          break;
        }
      }
      if(result) {
        return true;
      }
      arra1[j] = arra1[i];
      arra1[i] = temp;
    }
  }
  return false;
}

console.log(array_checking([10,20,30], [10,20,30]))
console.log(array_checking([10,20,30], [30,10,20]))
console.log(array_checking([10,20,30,40], [10,30,20,40]))

Sample Output:

true
false
true

Flowchart:

Flowchart: JavaScript - Check whether two arrays of integers and same length are similar or not

ES6 Version:

function array_checking(arra1, arra2) {

  for(let i = 0; i < arra1.length; i++) {
    for(let j = i; j < arra1.length; j++) {
      let result = true;
      const temp = arra1[i];
      arra1[i] = arra1[j];
      arra1[j] = temp;
      for(let k = 0; k < arra1.length; k++) {
        if(arra1[k] !== arra2[k]) {
          result = false;
          break;
        }
      }
      if(result) {
        return true;
      }
      arra1[j] = arra1[i];
      arra1[i] = temp;
    }
  }
  return false;
}

console.log(array_checking([10,20,30], [10,20,30]))
console.log(array_checking([10,20,30], [30,10,20]))
console.log(array_checking([10,20,30,40], [10,30,20,40]))

Live Demo:

See the Pen javascript-basic-exercise-87 by w3resource (@w3resource) on CodePen.


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Previous: Write a JavaScript program to find the types of a given array.
Next: Write a JavaScript program to check whether two given integers are similar or not, if a given divisor divides both integers and it does not divide either.

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JavaScript: Tips of the Day

Promise object

const myPromise = () => Promise.resolve('I have resolved!');

function firstFunction() {
  myPromise().then(res => console.log(res));
  console.log('second');
}

async function secondFunction() {
  console.log(await myPromise());
  console.log('second');
}

firstFunction();
secondFunction();

With a promise, we basically say I want to execute this function, but I'll put it aside for now while it's running since this might take a while. Only when a certain value is resolved (or rejected), and when the call stack is empty, I want to use this value.
We can get this value with both .then and the await keyword in an async function. Although we can get a promise's value with both .then and await, they work a bit differently.
In the firstFunction, we (sort of) put the myPromise function aside while it was running, but continued running the other code, which is console.log('second') in this case. Then, the function resolved with the string I have resolved, which then got logged after it saw that the callstack was empty.
With the await keyword in secondFunction, we literally pause the execution of an async function until the value has been resolved before moving to the next line.
This means that it waited for the myPromise to resolve with the value I have resolved, and only once that happened, we moved to the next line: second got logged.

Ref: https://bit.ly/3jFRBje