Use reduce() method to find a skipped number in an array of consecutive numbers

Question

I'm trying to use the reduce() method to find one skipped (missing) number in an array of (sometimes almost) consecutive numbers. Only one number will be missing at most. This is my codepen: http://codepen.io/PiotrBerebecki/pen/zBrRVd

For example,
findMissing([1,2,3,5]) should return 4

findMissing([1,2,3,4]) should return undefined

findMissing([2,3,4,6]) should return 5

findMissing([2,4,5,6]) should return 3

The code that I developed seems to work fine if there is indeed a number that was skipped. But it returns an undesired value if all numbers are present. Would you know how to fix it?

My JS code:

function findMissing(arr) {
  return arr.reduce(function(prev, curr) { 
    if (curr - prev !== 1) {
      return curr - 1;
    }
  });
}


// This should return 4, and it indeed returns 4
console.log(   findMissing([1,2,3,5])   );


// This should return 'undefined', but it returns 3
console.log(   findMissing([1,2,3,4])   );


// This should return 5, and it indeed returns 5
console.log(   findMissing([2,3,4,6])   );

UPDATE 1:

Based on the answers below, the following code delivers the desired outcome using the reduce() method:

// ****SOLUTION:****
function findMissing2(arr) {
  return arr.reduce(function(prev, curr, index, array) {
    if (curr === index + array[0]) {
      return prev;
    } else {
      return index + array[0]++;
    }
  }, void 0);
}

console.log(   findMissing2([1,2,3,4])   ); // Undefined
console.log(   findMissing2([1,2,3,5])   ); // 4
console.log(   findMissing3([2,3,4,6])   ); // 5
console.log(   findMissing2([2,3,4,5])   ); // Undefined
console.log(   findMissing2([2,4,5,6])   ); // 3

Answer 1

I would do this job as follows;

var a1 = [1,2,3,5],
    a2 = [2,3,4,5],
    a3 = [2,4,5,6],
  res1 = a1.reduce((p,c,i,a) => c == i+a[0] ? p : i + a[0]++, void 0),
  res2 = a2.reduce((p,c,i,a) => c == i+a[0] ? p : i + a[0]++, void 0),
  res3 = a3.reduce((p,c,i,a) => c == i+a[0] ? p : i + a[0]++, void 0);
console.log(res1);
console.log(res2);
console.log(res3);

Note: void 0 is a very safe undefined value in JS. The above code will mutate the tested array. You might prefer to call like a1.slice().reduce... if you want to keep the tested array as it is.