# AC O(n) Java Solution. Pretty Straightforward.

• ``````public boolean canPlaceFlowers(int[] flowerbed, int numberToPlace) {

int i=0;

if(numberToPlace == 0
|| (flowerbed.length == 1 && flowerbed[0] == 0 && numberToPlace == 1)) {
return true;
}

while(i < flowerbed.length) {
// if the current plot is planted, increment by 2.
if(flowerbed[i] == 1) {
i += 2;
continue;
}
// if the current plot is at 0 index and not planted and the next plot is empty.
else if(flowerbed[i] != 1 && i == 0) {
if(flowerbed[i+1] != 1) {
flowerbed[i] = 1;
numberToPlace--;
}
i += 1;
}
// if the current plot is last index and not planted and the previous plot is empty.
else if(flowerbed[i] != 1 && i == flowerbed.length-1) {
if(flowerbed[i-1] != 1) {
flowerbed[i] = 1;
numberToPlace--;
}
i += 1;
}
// if the current plot is not planted and prev and next plot is empty.
else if(flowerbed[i] != 1) {
if(flowerbed[i-1] != 1 && flowerbed[i+1] != 1) {
flowerbed[i] = 1;
numberToPlace--;
}
i += 1;
}
if(0 == numberToPlace) return true;
}
return false;
}
``````

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