Suppose that X is the array

X=[1  0  1  2  0  4  2  0  4  1  4  1]

We can compute the histogram with
H=HISTOGRAM(X,MIN=0,MAX=4,REVERSE_INDICES=R)

The min and max were set to the range of the data. Now print H
H=[3  4  2  0  3]
   0  1  2  3  4  Index

The index is printed here for reference. 
The vector H tells us that: 0 occurs 3 times since H[0]=3, 1 occurs 4 times, 
since H[1]=4, and so on. 

The locations of the values for each cell are given in the reverse index
vector, R. 

R=[ 6  9 13 15 15 18  1  4  7  0  2  9 11  3  6  5  8 10]
    0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17  Index

R[0]=6 tells us that the indexes for the zeros in the original X are 
listed beginning with position 6. 

R[1]=9 tells us that the indexes for the ones in the original X are 
listed beginning with position 9. 

R[2]=13 tells us that the indexes for the twos in the original X are 
listed beginning with position 13.

R[3]=15 tells us that the indexes for the threes in the original X are 
listed beginning with position 15.  

R[4]=15 tells us that the indexes for the fours in the original X are 
listed beginning with position 15.  

R[5]=18 tells us that the indexes for the fives in the original X are 
listed beginning with position 18.  Since 5 is greater than MAX, this
is a dummy entry to help with the following calculations.

If we were to print R[6:8] we would get the locations of the zeros in the X array.
They are R[6:8]=[1,4,7]. Hence, Print,X[R[6:8]] produces the output [1,4,7].

But we need a way to fill in the "6" and the "8" automatically. We could
do the following:

i=0
k1=R[i]        ;Yields k1=R[0]=6
k2=R[i+1]-1    ;Yields k2=R[1]-1=9-1=8
k3=R[k1:k2]    ;Yields k3=R[6:8]=[1,4,7]
Print,X[k3] ;Prints X[1,4,7]=[0,0,0]

Hence, k3 contains the indexes you want for the locations of 0 in the X array.

You can shorten this by collapsing some of the above statements:
i=0
k3=R[R[i]:R[i+1]-1]  ; The locations of 0 in X.

To find the location of 1 in X we write:
i=1
k3=R[R[i]:R[i+1]-1]

You can do this for each value using i=0,1,2,3,4.

Note that with i=4 the above gives
i=4
k3=R[R[i]:R[i+1]-1]=R[R[4]:R[5]-1]=R[15:18-1]=R[15:17]=[5,8,10]

We need R[5] in this formula, and the REVERSE_INDICES keyword provides it.

You can use the indices that are returned in this way to find the locations
of the values in any array.