## AMD Interview Question for Students

Country: India
Interview Type: Written Test

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

Stating this as optimization problem: Find max non decreasing sequence so that each species contributes at least one element. If max length is equal to length of species, answer to original question is TRUE.
Trying to solve this as simple DP might yield false positive results when all elements are taken from only one of the species, e.g. {{2,2,3,3},{4,1,1,1}}, to avoid this we will induce fake obstacles, a different pair of indexes from species A an B. This will force DP to make a switch at least once.
Overall complexity will be O(n^3).

``````private static final int INVALID = 1011;

public static boolean isSimilar(int[] a, int[] b) {
int len = a.length;
for (int i = 0; i<len; i++) {
int atmp = a[i];
a[i] = INVALID;
for (int j = 0; j < len; j++) {
if (i==j) { //no point exploring this setup since
//it will definitely not produce positive result
continue;
}
int btmp = b[j];
b[j] = INVALID;
if (processSim(a, b) == true) {
return true;
}
b[j] = btmp;
}
a[i] = atmp;
}

return false;
}

private static boolean processSim(int[] a, int[] b) {
int len = a.length;

int[][] maxLenMemo = new int[len+1][2];

maxLenMemo[0][0] = 0;
maxLenMemo[0][1] = 0;
maxLenMemo[1][0] = 1;
maxLenMemo[1][1] = 1;

for (int i = 2; i <= len; i++) {

int maxLen = 0;

if (a[i-1] != INVALID) {
if (a[i-1] >= a[i-2])
maxLen=(maxLen<maxLenMemo[i-1][0] + 1)?(maxLenMemo[i-1][0] + 1):maxLen;

if ( a[i-1] >= b[i-2])
maxLen=(maxLen<maxLenMemo[i-1][1] + 1)?(maxLenMemo[i-1][1] + 1):maxLen;

maxLenMemo[i][0] = maxLen;
}
maxLen=0;

if (b[i-1] != INVALID) {
if (b[i-1] >= b[i-2])
maxLen = (maxLen<maxLenMemo[i-1][1] + 1)?(maxLenMemo[i-1][1] + 1):maxLen;

if (b[i-1] >= a[i-2])
maxLen = (maxLen<maxLenMemo[i-1][0] + 1)?(maxLenMemo[i-1][0] + 1):maxLen;

maxLenMemo[i][1] = maxLen;
}
}

return Math.max(maxLenMemo[len][0], maxLenMemo[len][1]) == len;
}``````

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0
of 0 vote

c++, recursion, O(2 ^ n)

``````#include <iostream>
#include <vector>

using namespace std;

bool nonDecreasingSequence(vector<int>& a, vector<int>& b, vector<int>&c, int idx, int bit) {
if (idx == a.size()) {
return bit == 3;
}

if (idx == 0 || a[idx] >= c[idx - 1]) {
c[idx] = a[idx];
if (nonDecreasingSequence(a, b, c, idx + 1, bit | 1) == true) return true;
}

if (idx == 0 || b[idx] >= c[idx - 1]) {
c[idx] = b[idx];
if (nonDecreasingSequence(a, b, c, idx + 1, bit | 2) == true) return true;
}

return false;
}

int main(int argc, char* argv[]) {
vector<int> a, b, c;
int t, n, i;

cin >> t;
while (t--) {
cin >> n;
a.assign(n, 0);
b.assign(n, 0);
c.assign(n, 0);

for (i = 0; i < n; i++) cin >> a[i];
for (i = 0; i < n; i++) cin >> b[i];

if (nonDecreasingSequence(a, b, c, 0, 0) == true) {
for (i = 0; i < n; i++) {
cout << c[i] << " ";
}
cout << "\n";
} else {
cout << "IMPOSSIBLE\n";
}
}

return 0;
}``````

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0
of 0 vote

@kyduke, Have you considered a DP solution?

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0
of 0 vote

Greedy.

Name:

Writing Code? Surround your code with {{{ and }}} to preserve whitespace.

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