Google Interview Question for Java Developers


Country: United States




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public boolean find(int[][] a, int[][] b) {
        int start = b[0][0];
        for (int i = 0; i < a.length; i++) {
            for (int j = 0; j < a.length; j++) {
                if (a[i][j] == start && a.length - i >= b.length && a.length - j >= b.length) {
                    boolean res = matchArray(a, b, i, j);
                    if (res) return res;
                }
            }
        }
        return false;
    }

    public boolean matchArray(int[][] a, int[][] b, int i, int j) {
        int jOrig = j;
        for (int i1 = 0; i1 < b.length; i1++) {
            for (int j1 = 0; j1 < b.length; j1++) {
                if (b[i1][j1] != a[i][j++]) {
                    return false;
                }
            }
            i++;
            j = jOrig;
        }
        return true;
    }

- Anonymous February 12, 2018 | Flag Reply
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#include <iostream>
#include <vector>
//Author = Jean

bool isSubMatrix(const std::vector<std::vector<int>>& small, const std::vector<std::vector<int>>& large){
	int sizeYLarge = large.size();
	int sizeXLarge = large[0].size();
	int sizeYSmall = small.size();
	int sizeXSmall = small[0].size();
	int li,lj,si,sj;
	for (li = 0 ; li < sizeXLarge ; li++){
		for (lj = 0 ; lj < sizeYLarge ; lj++){
			if (large[li][lj] == small[0][0]){
				for (si = 0 ; si< sizeXSmall ; si ++){
					for (sj = 0 ; sj < sizeYSmall ; sj++) {
						std::cout<<"large "<<large[si+li][sj+lj]<<"small "<<small[si][li]<<std::endl;
						if (large[si+li][sj+lj]!=small[si][sj]){
							break;
						}
					}
				}
			}
		}
	}
	if (si == sizeXSmall && sj == sizeYSmall)
			return true;
	return false;
	}

main (){
	std::vector<std::vector<int>> large={{1 ,2 ,  3},
										 {4 ,5 ,  6},
										 {7 ,8 ,  9},
										 {10,11, 12}};

	std::vector<std::vector<int>> small={{5, 6},
										 {8,10}};
	std::cout<<isSubMatrix(small,large);

}

- Jean March 17, 2018 | Flag Reply
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of 0 vote

#include <iostream>
#include <vector>
//Author = Jean

bool isSubMatrix(const std::vector<std::vector<int>>& small, const std::vector<std::vector<int>>& large){
	int sizeYLarge = large.size();
	int sizeXLarge = large[0].size();
	int sizeYSmall = small.size();
	int sizeXSmall = small[0].size();
	int li,lj,si,sj;
	for (li = 0 ; li < sizeXLarge ; li++){
		for (lj = 0 ; lj < sizeYLarge ; lj++){
			if (large[li][lj] == small[0][0]){
				for (si = 0 ; si< sizeXSmall ; si ++){
					for (sj = 0 ; sj < sizeYSmall ; sj++) {
						std::cout<<"large "<<large[si+li][sj+lj]<<"small "<<small[si][li]<<std::endl;
						if (large[si+li][sj+lj]!=small[si][sj]){
							break;
						}
					}
				}
			}
		}
	}
	if (si == sizeXSmall && sj == sizeYSmall)
			return true;
	return false;
	}

main (){
	std::vector<std::vector<int>> large={{1 ,2 ,  3},
										 {4 ,5 ,  6},
										 {7 ,8 ,  9},
										 {10,11, 12}};

	std::vector<std::vector<int>> small={{5, 6},
										 {8,10}};
	std::cout<<isSubMatrix(small,large);

}

- Jean March 17, 2018 | Flag Reply
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Using kmp algorithm we can solve in n^3

- rahulkumarsk2015 May 30, 2018 | Flag Reply
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Here is O(n * m * k) or slightly faster than O(n^3) implementation using Finite Automata based on kmp, it is actually not very hard and described in the article "Pattern Searching | Set 6 (Efficient Construction of Finite Automata)".
O (n * m * k) space. It requires some space, but O(n^3) is faster than O(n^4) from the comments above.

But I expect that even O(n^3) isn't the ideal answer, I think the proper answer requires using Rabin-Karp algorithm, it is much harder to implement though.

var matrix = [
  [1,1,1,1,2],
  [1,1,1,1,2],
  [1,1,1,1,2],
  [1,1,1,2,2],
];
var smallMatrix = [
  [1,1],
  [1,2],
];
findMatrix(matrix, smallMatrix); // i: 2, j: 2

function findMatrix(matrix, smallMatrix) {
  var calculatedStates = [];
  
  for (var k = 0; k < smallMatrix.length; k++) {
    calculatedStates[k] = [];
    // build state machine for each row of small matrix
    var tf = buildTf(smallMatrix[k]);
    
    for (var i = 0; i < matrix.length; i++) {
      calculatedStates[k][i] = [];
      var state = 0;
      for (var j = 0; j < matrix[i].length; j++) {
        // state is how many numbers of smallMatrix[k]
        // are matched at (i,j)
        state = tf[state].get(matrix[i][j]) || 0;
        calculatedStates[k][i][j] = state;
      }
    }
  }
  
  for (var i = 0; i < matrix.length; i++) {
    for (var j = 0; j < matrix[i].length; j++) {
      var matchingRows =
          getMatchingRows(i, j, calculatedStates, smallMatrix);
      if (matchingRows == smallMatrix.length) {
        return {i: i, j: (j - smallMatrix[0].length + 1)};
      }
    }
  }
  
  return {i: -1, j: -1};
}

function getMatchingRows(i, j, calculatedStates, smallMatrix) {
  for (var k = 0; k < smallMatrix.length &&
       i + k < calculatedStates[k].length; k++) {
    if (calculatedStates[k][i + k][j] != smallMatrix[k].length) {
      return k;
    }
  }
  return smallMatrix.length;
}

function buildTf(nums) {
  var result = [];
  var lps = -1;
  for (var i = 0; i <= nums.length; i++) {
    result[i] = lps >= 0 ? new Map(result[lps]) : new Map();
    if (i < nums.length) {
      result[i].set(nums[i], i + 1);
      lps = lps >= 0 && result[lps].get(nums[i]) || 0;
    }
  }
  return result;
}

- inthecottonfield January 20, 2019 | Flag Reply


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