Interview Question


Country: United States




Comment hidden because of low score. Click to expand.
1
of 1 vote

function find (N, list) {
  var map = {}
  for (var i = 0; i < list.length; ++i) {
    if (list[i] <= N) {
      map[list[i]] = i
    }
  }
  console.log(map)
  closestD = Infinity
  closestPair = [null, null]
  for (var key in map) {
    if (({}).hasOwnProperty.call(map, key)) {
      var m = key
      var n = N - m
      var p = map[n]
      console.log('n,m', n,m)
      if ('' + n === m) {
       // Ignore things like the fact that 4 pairs with itself
       // to produce 4,4 d=0 ; a more robust implementation handles this
      } else if (p !== undefined) {
        var d = Math.abs(map[m] - p)
        console.log(m, p, N, n, d)
        if (d < closestD) {
          closestD = d
          closestPair = [m,n]
        }
      }
    }
  }
  console.log(closestD, closestPair)
}

find (8, [1,5,3,6,4,2])

- srterpe December 06, 2016 | Flag Reply
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0
of 0 vote

rivate static int[] getClosestSum(int target[], int sum, int diff, int index, int[] pair, int nextindex) {

		// int [] array = { 1, 5, 3, 6, 2};
		int[] nextPair = null;
		if (target.length == index || target.length == nextindex) {
			return pair;
		}

		nextPair = getClosestSum(target, sum, diff, index, pair, nextindex + 1);

		if (target[index] + target[nextindex] == sum) {

			int currentMin = target[index] - target[nextindex];
			int prevMin = pair[0] - pair[1];

			if (currentMin == Math.min(currentMin, prevMin) && currentMin >= 0) {
				pair[0] = target[index];
				pair[1] = target[nextindex];
			}

		}

		// nextPair = getClosestSum(target, sum, diff, index+1, pair,
		// nextindex+1);

		return pair;
	}

	public static void main(String[] args) {

		int[] array = { 1, 5, 3, 6, 2 };
		int[] pair = { Integer.MAX_VALUE, 0 };
		int[] result = null;
		int index = 0;
		for (int num : array) {
			result = getClosestSum(array, 8, 0, index++, pair, 1);
		}

		System.out.println(result);

		// given 8 = {5,3} and not {6,2}

	}

- MDX December 06, 2016 | Flag Reply
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0
of 0 vote

rivate static int[] getClosestSum(int target[], int sum, int diff, int index, int[] pair, int nextindex) {

		// int [] array = { 1, 5, 3, 6, 2};
		int[] nextPair = null;
		if (target.length == index || target.length == nextindex) {
			return pair;
		}

		nextPair = getClosestSum(target, sum, diff, index, pair, nextindex + 1);

		if (target[index] + target[nextindex] == sum) {

			int currentMin = target[index] - target[nextindex];
			int prevMin = pair[0] - pair[1];

			if (currentMin == Math.min(currentMin, prevMin) && currentMin >= 0) {
				pair[0] = target[index];
				pair[1] = target[nextindex];
			}

		}

		// nextPair = getClosestSum(target, sum, diff, index+1, pair,
		// nextindex+1);

		return pair;
	}

	public static void main(String[] args) {

		int[] array = { 1, 5, 3, 6, 2 };
		int[] pair = { Integer.MAX_VALUE, 0 };
		int[] result = null;
		int index = 0;
		for (int num : array) {
			result = getClosestSum(array, 8, 0, index++, pair, 1);
		}

		System.out.println(result);

		// given 8 = {5,3} and not {6,2}

	}

- manmad December 06, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

In ZoomBA:

/* 
 To solve this :
 1. Create a map with value as key, and index as value.
 2. Iterate the list with testing sum (S) S - item is in the map or not.
    2.1. If it is, 
       2.1.1 and the difference in position is 
       smaller than previous min store as min 
       2.1.2 Ignore 
    2.2. If it is not continue to [2] 
*/

def solution( arr, S ){
  // a map of key ( item ) -> item index 
  map = dict(arr) -> { [ $.o , $.i ] }
  min = [ -1,-1, num('inf') ] // store +infinity as minimum 
  fold ( arr , min ) -> {
    diff = S - $.o 
    if( diff @ map &&  map[diff] != $.i && $.p.2 > #|$.i - map[diff]| ){
      $.p = [ $.i , map[diff] , #|$.i - map[diff]| ]
    }
    $.p // returns 
  } 
}
arr = [1,4,5,3,6,4,2] 
#(i,j) = solution( arr , 8 )
printf( '%d,%d\n', arr[i], arr[j] )

- NoOne December 06, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

from itertools import combinations

def closet_sum(nums, n):
    print min([(abs(i-j), i, j) for (i,j) in combinations(nums,2) if i+j == n], key = lambda x: x[0])[1:]

- tsg December 12, 2016 | Flag Reply


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